Natural logarithms
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Natural logarithm with a calculator
Natural Logarithm with a Calculator
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 Use a calcuator to find log base e of 67 to the nearest thousandth
 So just as a reminder, e is one of these crazy numbers that shows up
 in natures and finance and all these things and it is approximately equal to 2.71 and it just keeps on and on
 so you could view log
 base is as 67 and you say "What does 'e' mean?" Well, e is just a number
 just like pi is just a number
 This is really the same thing as saying log base 2.71 and teh actual number
 so you have to write all the digits that just keep on going forever
 and never repeat of 67. So what power do I have to raise e to
 to get to 67? So another way of saying that is that if this is
 equal to x, e to the x is equal to 67, we need to figure out what x is
 Now, traditionally, you will never see people write log base e even though
 e is one of the most common bases to take a log of
 so the reason you wouldn't see log base e written this way
 is that log base e is the natural logarithm
 and I think that's used because e shows up so many times in nature.
 So log base e of 67, another way of saying that or seeing that and the more typical
 way of seeing that is the natural log. This is ln (I think this is from french or something)
 So this is the same thing as saying log base e
 of 67. This is saying that exact same thing.
 So what power do I have to raise e to to get 67. When you see this 'ln', is literally means
 'log base e'. Now, they let us use a calculator and that's good
 because I don't know off the top of my head what power you
 have to raise 2.71 and so forth to to get 67.
 So we'll get our calculator out. So we'll get our TI85 out.
 And different calculators will have different ways of doing it.
 If you have a graphing calulator like this, you can literally type in the
 statement, natural log of 67 and evaluate it.
 So here, this is the button for natural log, 'log naturale (fr),' maybe.
 Ln of 67, and then you press 'enter' and it will give you the answer.
 If you don't have a graphing calculator, you might have to press 67
 and then press 'ln' to give you the answer. But if you have
 a graphing calculator you can jsut type it in the same
 way as you would write it out.
 So 4.20469... and we want to roud to the nearest thousandth.
 So this is the thousandth place, this 4, the digit after that is 5
 or larger, it's a 6, so we are going to round up.
 So this is approximately equal to 4.205
 and it actualy makes a lot of sense, because we know that e is greater than 2
 and it is less than 3
 and if you think about 2 to the 4th power, that gets you to 16
 and 3 to the fourth power gets you to 81.
 67 is between 16 and 81, and e is between 2 and 3
 so at least if feels right
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?

Have something that's not a question about this content? 
This discussion area is not meant for answering homework questions.
If I am looking for a calculator to help me with this should I get a scientific calculator or graphing calculator?
I would say graphing. Either a Texas Instruments or a Casio, depending on what your school division uses (if applicable). If that's not a factor, Casios are more powerful, and some of the operations are more intuitive, but most instructions that I have found tend to be written for the TI series  and it's what Khan uses. Graphing calculators do everything that a scientific calculator does, and much more. Because of that, however, they tend to be a fair amount more expensive. If you plan on continuing in any math education, get the graphing calculator. Scientific calculators are just too limiting for higherlevel math.
The TI84 Plus Graphing Calculator is an excellent choice. I have one that allow you to change the base of an log without actually doing oddball math. It may need an update the the operating system, though.
I would have to agree with CasualJames. The biggest benefit that a graphing calculator has is the ability to create tables, and graphs which makes it easier to recheck your answers. I would recommend the TI84 plus, because it has many more operations that are much more userfriendly.
The TI series of graphing calculators are an excellent brand of calculators. I personally use the TI83plus, because it is cheaper than the TI84 and not a whole lot different than the newer version (besides being able to change the base of a log. If that's important to you, than consider the TI84plus, but its more $$$). As for scientific calculators, you should definitely have at least one because they are so much easier to operate than graphing calcs. Texas Instruments sells a variety of these simpler (but very useful) devices. For a final note, I wouldn't buy more than one graphing calculator  often computers can do the same things graphing calculators can, but even faster. Because of this, graphing calculators are more for students, whereas adults who need to do that sort of stuff usually use their laptops for the job. This reply might have come a bit late for jtfeliz, but I hope it shines some light on selecting a calculator that's right for you for anyone else who needs a calculator.
Where can I get the calculator Sal is using but for mac?
Both will be useful...but a scientific calculator is better in the long run....and its easier to use.
So does anyone know if he was right about that "log natural" french thing?
It's actually written "ln" instead of "nl" because the Latin name of natural log is "logarithmus naturali."
I was taught log natural by my high school teacher...
I was wondering the same too! I think he's wrong! But in french, we actually call this log "le logarithme népérien" and it's name after a Scottish mathematician John Neper working on logarithm tables, here's the wikipedia in english: https://en.wikipedia.org/wiki/John_Napier
..well I guess he's right in the fact that we place the term "logarithme" before "népérien", further abbreviated "ln" !
Logarithmus naturali is how I was taught it too, latin for natural log.
Where and how does "e" appear in nature?
e=(1+(1/x))^x The variable "e" is often used in calculating equations in physics, such as Newton's Law of Cooling (look it up, I won't explain it as well as others will). It is also used in finance with compound interest. Mr. Khan probably has something on it in his finance videos.
"e" appears in nature in the design of flowers, and the placement of designs on turtle shells and other such natural geometric shapes.
see the compound interest video, he explains e there
What if you have a question that asks like e^2x+1=55? How can you solve for e?
Ruaida,
"e" is a transcendental number, kind of like "pi." So you don't need to solve for it. We already know what 'e' is. It's about 2.71828182845….
What you probably want to solve for is "x". Here is how you would do that:
(e^2x) +1=55. (I added the parenthesis around e^2x just to make it a little more clear) Subtract 1 from both sides:
e^2x=54. Take the natural logarithm of both sides (that's 'ln' on your calculator usually, not 'log')
2x=ln 54. Divide both sides by 2.
x=(1/2)ln (54) (By the way, this is the same as x=ln (54^1/2) or ln (sqrt(54)).
x= 1/2 *3.98898… = 1.9945 (approximately).
"e" is a transcendental number, kind of like "pi." So you don't need to solve for it. We already know what 'e' is. It's about 2.71828182845….
What you probably want to solve for is "x". Here is how you would do that:
(e^2x) +1=55. (I added the parenthesis around e^2x just to make it a little more clear) Subtract 1 from both sides:
e^2x=54. Take the natural logarithm of both sides (that's 'ln' on your calculator usually, not 'log')
2x=ln 54. Divide both sides by 2.
x=(1/2)ln (54) (By the way, this is the same as x=ln (54^1/2) or ln (sqrt(54)).
x= 1/2 *3.98898… = 1.9945 (approximately).
do you mean solve for x? since e is already quantified like π, it'd be a bit strange to solve for e. If you meant x; follow this approach, take a ln of each side so that you get lne^2x + ln1= ln55
then use log rules, to get 2xlne+ln1=ln55, then solve for x. where x= (ln55ln1)/2... also just in case you wondered where lne went, its actually equal to 1. hope that helps a bit
then use log rules, to get 2xlne+ln1=ln55, then solve for x. where x= (ln55ln1)/2... also just in case you wondered where lne went, its actually equal to 1. hope that helps a bit
Does anyone know how to change the base on a TI83 to something other than ten?
actually ON ALL CALCULATORS you have to do this equation, log a(b) = log b(b)/log b(a)
You cant sorry your calculator doesnt have it programed but give me problem your having trouble with i can help. =)
Thank you CasualJames I was trying to find the formula, but with limited resources that is a little difficult.
Hey Lenin,
if you want to know what log 1 (2)=
you press this on the calc: log(1)/log(2) and voila! you have your answer
if you want to know what log 1 (2)=
you press this on the calc: log(1)/log(2) and voila! you have your answer
To echo the previous answers, you can't on an 83.
The change of base formula, incidentally, is log a(b) = log b(b)/log b(a)
The change of base formula, incidentally, is log a(b) = log b(b)/log b(a)
Sal made another video on how to calculate different base logarithms with a basic calculator. Log base(x) of (y) equals log(y)/log(x).
Ryan Farias is right TI83 cannot change the base without doing a different formula. I am confused as to what that formula is though... I know that it has something to do with dividing the base and/or the log by one or the other.
I'm not sure about TI83, but on TI84 plus, I know this works:
Press MATH
Scroll down the MATH list until to see logBASE(
Press ENTER
This will give you the LOG(base A of B) function.
Let me know if it works on TI83! Hope it helps! =)
Press MATH
Scroll down the MATH list until to see logBASE(
Press ENTER
This will give you the LOG(base A of B) function.
Let me know if it works on TI83! Hope it helps! =)
What if you have a number in front of the e instead of log? Like...4e^x = 10? How would you go about doing that?
Taylor,
Try dividing both sides by 4 first.
e^x = 10/4
Now can you solve?
Try dividing both sides by 4 first.
e^x = 10/4
Now can you solve?
Yeah. You would divide 10 and 4, and get 2.5, and then the final answer would be x = 0.92 (rounded to the hundredth place), right?
What about if you had e^x4 = 2? Or any other problem that adds to the x, divides the x by a number, or does e^ln 5x? I have a test on Friday on all of this stuff, and I'm completely clueless.
What about if you had e^x4 = 2? Or any other problem that adds to the x, divides the x by a number, or does e^ln 5x? I have a test on Friday on all of this stuff, and I'm completely clueless.
For problems that add/subtract to/from the x, simply solve for the exponent by using ln. In the example you gave:
e^(x4) = 2
x  4 = ln(2)
x = ln(2) + 4
An example for division:
e^(x/5) = 2
Same thing as before.
Use the ln.
x/5 = ln(2)
x = 5 ln(2)
For your last example let's equate it to some constant just for the sake of clarity. We'll choose 2 because it's a really friendly number:
e^(ln(5x)) = 2
Now, we have an important identity for logs. x^(log(N) base x) = N
so e^(ln(5x)) = 5x
so 5x = 2 and finally x = 2/5 Hope this helps. :)
e^(x4) = 2
x  4 = ln(2)
x = ln(2) + 4
An example for division:
e^(x/5) = 2
Same thing as before.
Use the ln.
x/5 = ln(2)
x = 5 ln(2)
For your last example let's equate it to some constant just for the sake of clarity. We'll choose 2 because it's a really friendly number:
e^(ln(5x)) = 2
Now, we have an important identity for logs. x^(log(N) base x) = N
so e^(ln(5x)) = 5x
so 5x = 2 and finally x = 2/5 Hope this helps. :)
how do you go on solving problems like this 4^−9t = 0.60 ... it's not an homework question I am just puzzled what ln is or if it's a problem like this where do you plug it in?
ln, the natural logarithm, is log base e ( an irrational number, not a variable). It is important in calculus. I do not think that it would help you with 4^9t = 0.6, though. Here is how:
4^9t = 0.6
9t*log(4) = log(0.6)
t*(9*log(4)) = log(0.6)
t= log(0.6)/(9*log(4))
4^9t = 0.6
9t*log(4) = log(0.6)
t*(9*log(4)) = log(0.6)
t= log(0.6)/(9*log(4))
When you have the variable you're trying to solve in an exponent, you want to "bring it down to normal level". As you point out, ln can help you a lot:
4^−9t = 0.60
ln (4^(−9t)) = ln 0.60 [take the natural log of both sides]
9t ln 4 = ln 0.60 [use log property logₓyⁿ = nlogₓy {x is the base which in our case is e}]
ln 4 and ln 0.60 are just constants so solving for t should now be straightforward:
9t = (0.5108...)/(1.3862...) = 0.3684...
=> t = 0.0409...
Wolfram Alpha confirms our work: http://www.wolframalpha.com/input/?i=4%5E%28%E2%88%929t%29+%3D+0.60
4^−9t = 0.60
ln (4^(−9t)) = ln 0.60 [take the natural log of both sides]
9t ln 4 = ln 0.60 [use log property logₓyⁿ = nlogₓy {x is the base which in our case is e}]
ln 4 and ln 0.60 are just constants so solving for t should now be straightforward:
9t = (0.5108...)/(1.3862...) = 0.3684...
=> t = 0.0409...
Wolfram Alpha confirms our work: http://www.wolframalpha.com/input/?i=4%5E%28%E2%88%929t%29+%3D+0.60
Where does "e" come up in nature? Just curious.
"e" is the natural representation for any problem involving exponential growth. For example, halflife problems are typically expressed at the college level using "e", as it gives you a clean connection between the amount of the radioactive substance remaining and the current rate of decay (the level of radiation).
Ln actually mean "Logarithm Népérien" reffering to John Napier, who discovered Logarithms. http://en.wikipedia.org/wiki/John_Napier
So I have this assignment where I need to condense an expression with two natural logarithms...
2 ln 7  3 ln 4.... I don't get it. At all.
2 ln 7  3 ln 4.... I don't get it. At all.
It should work the same way it does with log. You could put the multipliers (2 for the first term, 3 in the second) within the log operation as an exponent of the quantity inside the logarithm.
If you start with a*log(x) that equals log(x^a) as demonstrated in the logarithm set of videos.
Another property is that log(x)log(y)=log(x/y), also demonstrated in the previous section.
Since ln is just log at base e instead of base 10, it responds to the properties identically.
If you start with a*log(x) that equals log(x^a) as demonstrated in the logarithm set of videos.
Another property is that log(x)log(y)=log(x/y), also demonstrated in the previous section.
Since ln is just log at base e instead of base 10, it responds to the properties identically.
What is an mathematical explation of what e stands for?
No, e has several definitions. Here is 1 
e = limit as n approaches infinity of (1 +1/n)^n
Go to the precalculus playlist, and watch the videos on compound interest and e
e = limit as n approaches infinity of (1 +1/n)^n
Go to the precalculus playlist, and watch the videos on compound interest and e
ACTUALLY ↑←hkapur97→↓ e^iπ =1 and e=lim_x→∞(1+1/x)^x
If you have a dollar and you have an interest rate of 100% that is compounded an infinite amount of times every second then at the end of the year you will have e dollars which is about 2.71828183
e^x is f(x) when (f´(x))/(f(x)=1
not sure if that was that helpful
not sure if that was that helpful
look it up on Wikipedia, you'll get a really good definition there.
so that means the iπ√1=e and log_e〖1〗=iπ
I was wondering, when considering the expression ln(2e)^2, does the the square belong to the "ln" function as a whole or does it belong to the "2e" separately?
To the whole: `(ln(2e))^2`
I didn't see that formula in the video. I would have to see the original context to be sure something wasn't lost in the translation to text.
I didn't see that formula in the video. I would have to see the original context to be sure something wasn't lost in the translation to text.
To the logarithm's argument. You would write that squared logarithm as `ln^2 (2e)`.
Thanks so much guys :), this makes a huge difference. The original question was, " Simplify ln(2e)^2 + ln((4e^6)^0.5)  ln(8e^7) without the use of a calculator."
You can simplify this using the property of logs:
(A) `ln(xy) = ln(x) + ln(y)`
(B) `ln(x^a) = a*ln(x)`
This is covered in the "Logarithm basics" section, which you may want to review. (Note that B can be derived from A, but it happens often enough to memorize.)
Looking at the whole formula, I think the parenthesis are misplaced. It should be:
`ln(2e^2)`
which expands to:
`ln(2)+ln(e)+ln(e)`
or
`ln(2)+2*ln(e)`
The other terms can be expanded in the same way. Note that 4 and 8 are powers of 2. The 1/2 exponent can be simplified using (B), or by using the properties of exponents. If you work it both ways, you will see why logs are cool.
Please keep asking if you need more help.
(A) `ln(xy) = ln(x) + ln(y)`
(B) `ln(x^a) = a*ln(x)`
This is covered in the "Logarithm basics" section, which you may want to review. (Note that B can be derived from A, but it happens often enough to memorize.)
Looking at the whole formula, I think the parenthesis are misplaced. It should be:
`ln(2e^2)`
which expands to:
`ln(2)+ln(e)+ln(e)`
or
`ln(2)+2*ln(e)`
The other terms can be expanded in the same way. Note that 4 and 8 are powers of 2. The 1/2 exponent can be simplified using (B), or by using the properties of exponents. If you work it both ways, you will see why logs are cool.
Please keep asking if you need more help.
What I'm more interested in is how to calculate this WITHOUT a calculator
To compute a logarithm (except for simple cases where it is an integer or easy fraction) without a calculator or a table of logs, is very difficult. I can show you the calculation, though:
ln(x) = ∑(n=1 to ∞) 2*[(x1)/(x+1))]^(2n1)/(2n1)
Doing this for ln(4) we get:
ln (4) = 6/5 + 18/125 + 486/15625 + 4374/546875 + 4374/1953125 ....
The first 5 elements of the series sum to about 1.38534...
The actual ln (4) = 1.38629...
To get very many digits correct, you would need to add the first 20 or 30 members of the series.
ln(x) = ∑(n=1 to ∞) 2*[(x1)/(x+1))]^(2n1)/(2n1)
Doing this for ln(4) we get:
ln (4) = 6/5 + 18/125 + 486/15625 + 4374/546875 + 4374/1953125 ....
The first 5 elements of the series sum to about 1.38534...
The actual ln (4) = 1.38629...
To get very many digits correct, you would need to add the first 20 or 30 members of the series.
Unfortunately, it's not very practical. Unless you have somethin really simple, like ln(e^3) = 3, It's impossible to work out a logarithm without a calculator, unless you want to spend hours doing trial and error.
can someone help me with this equation please
2ln x + ln 5 = 7.2
2ln x + ln 5 = 7.2
How do you make the radical or square root sign on your computer?
it will become
ln x^2(5)= 7.2
e^7.2=x^2(5)
1339.4308=x^2(5)
1339.4308/5=x^2(5)/5
√267.8862=√x^2
16.3672=x
ln x^2(5)= 7.2
e^7.2=x^2(5)
1339.4308=x^2(5)
1339.4308/5=x^2(5)/5
√267.8862=√x^2
16.3672=x
can anyone help me with this equation
e^x5e^x=4
e^x5e^x=4
e^x5*e^(x)=4>e^x5*(1/(e^x))=4>e^x5/(e^x)=4. Let e^x be n.
n5/n=4 (Multiply each side by n)>n^2+5=4n (Subtract each side by 4n)>n^24n+5=0 (Factor)>(n5)(n+1)=0.
n5=0 or n+1=0, thus, n=5 or n=1. Now, we substitute e^x for n: e^x=5 or e^x=1. Take the natural log of each side: ln(e^x)=ln(5) or ln(e^x)=ln(1) (Simplify)>x=ln(5) or x=ln(1).
The natural log of a number is the number of times you would have to multiply e to get that number. However, no matter how many times you multiply e, you can't get a negative number. So, our answer is ln(5), which is approximately 1.60943791.
n5/n=4 (Multiply each side by n)>n^2+5=4n (Subtract each side by 4n)>n^24n+5=0 (Factor)>(n5)(n+1)=0.
n5=0 or n+1=0, thus, n=5 or n=1. Now, we substitute e^x for n: e^x=5 or e^x=1. Take the natural log of each side: ln(e^x)=ln(5) or ln(e^x)=ln(1) (Simplify)>x=ln(5) or x=ln(1).
The natural log of a number is the number of times you would have to multiply e to get that number. However, no matter how many times you multiply e, you can't get a negative number. So, our answer is ln(5), which is approximately 1.60943791.
How would you solve y =ln( x3/5  8) for x?
Start with the outermost functions, and 'undo' them on each side. The 'opposite' of the ln function is the exp function, so we have exp[y] = exp[ln(x^3 / 5  8)], and therefore exp[y] = x^3 / 5  8. Now add 8 to both sides: exp[y] + 8 = x^3 / 5. Multiply both sides by 5: 5exp[y] + 40 = x^3. Now cuberoot both sides: (5exp[y] + 40)^(1/3) = x, and you've solved for x.
Thankx! but somehow ur answer isnt the same of my teachers! This is how he solved it:
y =ln(x3/58)
e^y=x^3/58
e^y+8=x^3/5
(e^y+8)^5/3=x
x=(e^y+8)^5/3
Could u explain this?
y =ln(x3/58)
e^y=x^3/58
e^y+8=x^3/5
(e^y+8)^5/3=x
x=(e^y+8)^5/3
Could u explain this?
I've heard something about mantessa and characteristic of logarithm...can i find any video on that?
That was a practical part of calculating (base 10) logarithms back before scientific calculators existed. But it isn't something that modern students will ever really need to know, so I wouldn't hold my breath waiting for Sal to make a video about it.
In a nutshell, people used to find logarithms using tables, but the tables would only give the logarithms of numbers between 1 and 9.9999. So, if you wanted to calculate log 5280, you'd say log 5280 = log (1000 * 5.28) = log 1000 + log 5.28 = 3 + log 5.28 = 3.7226. (You'd be able to look up that last bit on the table since it's between 1 and 10.) So you can see how the integer and fractional part of that logarithm are separate keys to finding out what the number is, and the fancy mathematical words for the parts 3 and 0.7226 of that logarithm are the characteristic and mantissa of the logarithm. Of course, nowadays you just push one button on a calculator and it does all that work for you, so the concept doesn't impact much of anyone any more.
In a nutshell, people used to find logarithms using tables, but the tables would only give the logarithms of numbers between 1 and 9.9999. So, if you wanted to calculate log 5280, you'd say log 5280 = log (1000 * 5.28) = log 1000 + log 5.28 = 3 + log 5.28 = 3.7226. (You'd be able to look up that last bit on the table since it's between 1 and 10.) So you can see how the integer and fractional part of that logarithm are separate keys to finding out what the number is, and the fancy mathematical words for the parts 3 and 0.7226 of that logarithm are the characteristic and mantissa of the logarithm. Of course, nowadays you just push one button on a calculator and it does all that work for you, so the concept doesn't impact much of anyone any more.
But in our exams we are not allowed to use scientific calculators..and I can't find any sense in the procedures dictated by my tutor about searching natural sines ,cosines, tangents and all that stuff in a log book...(Clarke's Table)..I really wish Sal made a video..I don't know where else I must post this!
Can someone prove this for me?
lim (a^x 1)/x = ln(a)
x>0
Thanks. :)
lim (a^x 1)/x = ln(a)
x>0
Thanks. :)
Let me know if you want this proved, but the derivative of aᵡ = aᵡ · ln(a) for a > 0.
First, let's see what happens if we just plug x=0 into our expression:
(a⁰1) / 0 = (11) / 0 = 0 / 0. This is undefined and we can use L'Hôpital's rule (check out the link if this is unfamiliar to you). L'Hôpital's rule states that if a limit evaluates to 0/0 (or (±∞)/(±∞), I believe), you can evaluate the limit of the derivative of the nominator divided by the derivative of the denominator (differentiate them both independently). Let's try that:
1. Differentiate both the numerator and the denominator of our expression:
• d/dx [aᵡ  1] = aᵡ · ln(a)  0 = aᵡ · ln(a)
• d/dx [x] = 1
2. Rewrite the original expression as (aᵡ  1)' / x':
(aᵡ · ln(a)) / 1 = aᵡ · ln(a)
3. Reevaluate the limit (let's plug in x=0 once again. If you still end up with 0/0 or (±∞)/(±∞), you can keep on using the rule until you eventually end up with something else):
Plug in x=0:
a⁰ · ln(a) = 1 · ln(a) = ln(a)
We have now proved that the limit of (aᵡ  1) / x = ln(a) as x approaches 0.
An introduction to L'Hôpital's rule:
https://www.khanacademy.org/math/calculus/derivative_applications/lhopital_rule/v/introductiontolhopitalsrule
First, let's see what happens if we just plug x=0 into our expression:
(a⁰1) / 0 = (11) / 0 = 0 / 0. This is undefined and we can use L'Hôpital's rule (check out the link if this is unfamiliar to you). L'Hôpital's rule states that if a limit evaluates to 0/0 (or (±∞)/(±∞), I believe), you can evaluate the limit of the derivative of the nominator divided by the derivative of the denominator (differentiate them both independently). Let's try that:
1. Differentiate both the numerator and the denominator of our expression:
• d/dx [aᵡ  1] = aᵡ · ln(a)  0 = aᵡ · ln(a)
• d/dx [x] = 1
2. Rewrite the original expression as (aᵡ  1)' / x':
(aᵡ · ln(a)) / 1 = aᵡ · ln(a)
3. Reevaluate the limit (let's plug in x=0 once again. If you still end up with 0/0 or (±∞)/(±∞), you can keep on using the rule until you eventually end up with something else):
Plug in x=0:
a⁰ · ln(a) = 1 · ln(a) = ln(a)
We have now proved that the limit of (aᵡ  1) / x = ln(a) as x approaches 0.
An introduction to L'Hôpital's rule:
https://www.khanacademy.org/math/calculus/derivative_applications/lhopital_rule/v/introductiontolhopitalsrule
Here in the question in the above video , what does that nearest thousand means ?
Just to add to the already good answer above by Ammara Khan,, a thousandth is 3 decimals. How come? Well, one tenth is 1 decimal, because one tenth is 0,1. Maybe this is what you mean with your question,
where and how does e appear in nature
Is there any video that addresses solving or simplifying natural logarithm problems WITHOUT a calculator?
Watch the videos on logarithm properties, and the two video in this play list https://www.khanacademy.org/math/trigonometry/exponential_and_logarithmic_func/log_functions/v/solvinglogarithmicequations_DUP_1
How do you Simplify ln(2e)^2 + ln((4e^6)^0.5)  ln(8e^7) without the use of a calculator?
I need help finding a natural log with a variable in the exponent. For example, 5e^(.035x)=200, which reduces to e^(.035x)=40 because I don't know where to go from there.
e^(.035x)=40 From this step, you can manipulate a little bit. So it becomes:
log(base e)40=0.035x
Then go on to solve the problem.
For your example, 0.035x=3.688879(correct to 6 decimal places)
x=105.39656(correct to 5 d.p.)
After you get an answer, try to plug it in. If it is close to the original number, then you have got it right.
Hope this helps :)
log(base e)40=0.035x
Then go on to solve the problem.
For your example, 0.035x=3.688879(correct to 6 decimal places)
x=105.39656(correct to 5 d.p.)
After you get an answer, try to plug it in. If it is close to the original number, then you have got it right.
Hope this helps :)
How do you find 3.03=1.086 to the power x without using log?
3.03=1.086^x? I'm afraid you have to use logs for that. Those are the equations when logarithms show their real power :)
If you take log base 1.086 to both sides, you get x= log_1.086 (3.03). Of course, you're not expected to calculate something like that, so if you need a decimal approximation, it's the calculator's job.
If you take log base 1.086 to both sides, you get x= log_1.086 (3.03). Of course, you're not expected to calculate something like that, so if you need a decimal approximation, it's the calculator's job.
How would I solve (e^x)^5=1000?
first what exactly is it asking you to solve for e, x, ?
e^5x = 1000 (rule of exponents: power to a power)
=> e^5x/5 = 1000^1/5 (take the 5th root)
e^x = 1000^1/5 (which is approximately 3.98, if rounded to hundredths)
this may get you a little further..
e^5x = 1000 (rule of exponents: power to a power)
=> e^5x/5 = 1000^1/5 (take the 5th root)
e^x = 1000^1/5 (which is approximately 3.98, if rounded to hundredths)
this may get you a little further..
What if e is raised to a power like ee^(2x)=1?
When do you use natural log and when do you use regular logarithms? I mean, what's the point of using a log with base e when you could just use something simple like log with base 10?
The log is the inverse of the exponential function. So if you had a problem the had say e^x=1 you could simplify this problem by taking the natural log of both sides since it is e to a power and the natural log is log base e.
I'm still slightly confused  I don't really understand the significance of the number "e." For example, pi is the ratio between the diameter and circumference of a circle, it's used in the Golden Ratio, etc. What's "e" used for, and how did we figure out what its value is?
Thank you both!
Like π, e comes up again and again throughout both pure mathematics and the mathematics of the real world. In fact, in my own profession (Chemistry, though I'm retired now) we run into e far more than we run into π.
e, like π, is both irrational and transcendental, so we cannot write down the exact value of e. It is roughly 2.71828183...
e, like π, is both irrational and transcendental, so we cannot write down the exact value of e. It is roughly 2.71828183...
e is Euler's constant. The number e is of importance in mathematics, alongside 0, 1, π and i. All five of these numbers play important and recurring roles across mathematics, and are the five constants appearing in one formulation of Euler's identity.
Here is a video on Euler's Formula and Euler's Identity that helps explain e.
http://www.khanacademy.org/math/calculus/sequences_series_approx_calc/maclaurin_taylor/v/eulersformulaandeulersidentity
Here is a video on Euler's Formula and Euler's Identity that helps explain e.
http://www.khanacademy.org/math/calculus/sequences_series_approx_calc/maclaurin_taylor/v/eulersformulaandeulersidentity
so what does 2.h have to do with it all does that help get the answer or do we have to use the calculator
how do i solve the equation to find x 5=ln(x+1)
how do you solve
e^e^x = 3
e^e^x = 3
e^eˣ = 3
ln e^eˣ = ln 3
eˣ ln e = ln 3
eˣ = ln 3
ln eˣ = ln(ln 3)
x ln e = ln(ln 3)
x = ln(ln 3)
x ≈ 0.094048
ln e^eˣ = ln 3
eˣ ln e = ln 3
eˣ = ln 3
ln eˣ = ln(ln 3)
x ln e = ln(ln 3)
x = ln(ln 3)
x ≈ 0.094048
which videos cover multiple/submultiple angles in trigonometry
check algebra.
thanks ill check
how would you find Ln 3x=2? thank and please show me steps to do it.
e^0.06t/(1+(0.06/365)^365t = 1000/900
ln is latin for logarithm naturalus
How to find the value of a logarithm using a logarithm book. In India, we calculate logarithms using a log book. So does anyone know,how to calculate using the book?
How do you do natural logs without calculators? Where can I find videos to help me with collegelevel calculus?
From where can we get the TI85 which Khan Academy uses. If an emulator is there please tell me where i can download it from.
You can probably find it or something like it here: http://www.ticalc.org/programming/emulators/software.html
How did 'e' come into existence? Why use 'e' when you get an approximation?
e is the "natural exponential base". It is typically defined as the limit of: (1+1/n)^n as n approaches infinity (though I'll agree that isn't the most intuitive definition). Alternatively, the rate of increase of e^x at any point on the graph is equal to the value of the function itself. e^x defines continuous exponential growth at a 100% growth rate.
In calculus, the derivative of ln(x) = 1/x and the derivative of e^x = e^x. While you can also take the derivative of functions using other bases, these formulas are "clean" in that they don't need any additional constants to adjust. And using a Taylor series, e^x = 1 + x + x^2/2 + x^3/6 + .... + x^n/n! + .... Again, a "clean" formula.
It is a little like pi with circles. If a circle has an integer radius, then the circumference and area will be most easily (and exactly) expressed as a multiple of pi. Similarly, continuous exponential growth is most easily and exactly expressed using a base of e.
In calculus, the derivative of ln(x) = 1/x and the derivative of e^x = e^x. While you can also take the derivative of functions using other bases, these formulas are "clean" in that they don't need any additional constants to adjust. And using a Taylor series, e^x = 1 + x + x^2/2 + x^3/6 + .... + x^n/n! + .... Again, a "clean" formula.
It is a little like pi with circles. If a circle has an integer radius, then the circumference and area will be most easily (and exactly) expressed as a multiple of pi. Similarly, continuous exponential growth is most easily and exactly expressed using a base of e.
how do you graph log base 2 of (x + 4) on the ti83 plus calculator
log base 2 of (x+4) = (ln(x+4))/(ln(2)). Try typing that in. I know that ln() is on there somewhere.
how do i solve ln(5)1 ?
If you want the exact value, use a calculator.
I don't see anything much you can do with `ln(5)  1`.
I don't see anything much you can do with `ln(5)  1`.
can u solve this without a calculator?
if your a super genius and can solve decimal problem with only using your mind
To be Ln(xy)<0, the solution was e is greater than one so xy is less one. Can you explain me why? Thank you!
the expression e^x/e^(x+4) can be written as e^f(x), where f(x) is a function of x. find f(x).
I don't really understand the concept of eliminating ln and e together. In my book, an example says:" ln 5x = 4" and there's a step where it converts the problem to 5x = e^4. Can anyone explain it indepth on how the e^ln or ln e = 1 works?
Thanks a lot
Thanks a lot
In my other answer, I demonstrated some of the basic log mixed with exponent properties. Now I will solve the problem you mentioned, but explain each step.
ln 5x = 4
We need to solve for x, so we need to separate it out from either a log or an exponent.
As in demonstrated in my other answer a^(logₐ(x)) = x
Since ln(x) = logₑ(x), ln and e^ undo each other (as shown in my other answer) And,
Using the property that if a=b then nᵃ = nᵇ, and choosing the undo base (e) for the ln x we get:
ln 5x = 4
e^(ln(5x)) = e⁴
Using the undo property of samebase exponent and logarithm:
5x = e⁴
x = ⅕e⁴
Thus x is an irrational and transcendental number. It is approximately:
x ≅ 10.919630
ln 5x = 4
We need to solve for x, so we need to separate it out from either a log or an exponent.
As in demonstrated in my other answer a^(logₐ(x)) = x
Since ln(x) = logₑ(x), ln and e^ undo each other (as shown in my other answer) And,
Using the property that if a=b then nᵃ = nᵇ, and choosing the undo base (e) for the ln x we get:
ln 5x = 4
e^(ln(5x)) = e⁴
Using the undo property of samebase exponent and logarithm:
5x = e⁴
x = ⅕e⁴
Thus x is an irrational and transcendental number. It is approximately:
x ≅ 10.919630
First, understand this:
If a = b, then xᵃ = xᵇ where x is any number.
Now understand that a log and an exponent with the same base undo each other, thus:
logₐ(aˣ) = x
and a^(logₐ(x)) = x
I can go through the proof of this property if it helps, but let me just show you an informal reason:
logₐ(aˣ) = xlogₐ(a)
Since logₐ(n) is asking us what power does a have to be raised to in order to equal n, then if they are the same number, the power is always 1, because a¹=a. Thus, logₐ(a) = 1 (provided a is an allowed number for a logarithm base).
Therefore logₐ(aˣ) = xlogₐ(a) = (x)(1) = x,
Showing that a^(logₐ(x)) = x is a bit more tedious, so I will just show that the equation is true rather than a more direct proof:
a^(logₐ(x)) = x
Take the logₐ of both sides:
logₐ[a^(logₐ(x))] = logₐ(x)
Use the property that log (nᵇ) = (b)log(n)
logₐ(x)(logₐa) = logₐ(x)
Use the property that logₓ(x) = 1 (see above)
logₐ(x)(1) = logₐ(x)
logₐ(x) = logₐ(x)
By law of identity, since both logs have the same base, it must be the case that:
x = x
Thus, the equation is true that a^(logₐ(x)) = x
Since ln(x) means logₑ(x)
Thus ln(eˣ) = x and e^(ln(x)) = x
If a = b, then xᵃ = xᵇ where x is any number.
Now understand that a log and an exponent with the same base undo each other, thus:
logₐ(aˣ) = x
and a^(logₐ(x)) = x
I can go through the proof of this property if it helps, but let me just show you an informal reason:
logₐ(aˣ) = xlogₐ(a)
Since logₐ(n) is asking us what power does a have to be raised to in order to equal n, then if they are the same number, the power is always 1, because a¹=a. Thus, logₐ(a) = 1 (provided a is an allowed number for a logarithm base).
Therefore logₐ(aˣ) = xlogₐ(a) = (x)(1) = x,
Showing that a^(logₐ(x)) = x is a bit more tedious, so I will just show that the equation is true rather than a more direct proof:
a^(logₐ(x)) = x
Take the logₐ of both sides:
logₐ[a^(logₐ(x))] = logₐ(x)
Use the property that log (nᵇ) = (b)log(n)
logₐ(x)(logₐa) = logₐ(x)
Use the property that logₓ(x) = 1 (see above)
logₐ(x)(1) = logₐ(x)
logₐ(x) = logₐ(x)
By law of identity, since both logs have the same base, it must be the case that:
x = x
Thus, the equation is true that a^(logₐ(x)) = x
Since ln(x) means logₑ(x)
Thus ln(eˣ) = x and e^(ln(x)) = x
why cant you just us log e of n instead of lnx
It's simply because log e is used so much that having a shorter abbreviation makes sense. Mathematically, they are the same. It's just a naming convention.
At 5:31,what is mantisa?
Due to Euler's identity does that not mean that the Ln of any algebraic number is transcendental as e^x, if x is algebraic, equals a transcendental number? or did i misunderstand Euler's identity?
Both e^x and ln(x) are transcendental if x is algebraic.
how do we simplify expressions such as 2^2x4^3x/16x?
how would you solve a natural log equation without a calculator?? for ex:
ln(x)  ln(4) = 7
is there a video on this already?
ln(x)  ln(4) = 7
is there a video on this already?
How long will I t take for $4,000 to grow to $17,000 if the money is invested at 7.7% compounded quarterly?( use the natural logarithm, ln, for this question's solution).
The square root of 69 is 8 something. Right? 'Cause I've been tryin to work it out all....(I still don't get what it means)
(It is a math question, right?)
(It is a math question, right?)
if you use a calculator, it says that sqrt69=8.306623863
Where does e show up in nature?
e doesn't show up in nature. In calculus it is the iπ root of 1 because e^iπ=1 so log 1=iπ
e
e
How much is the TI 85 calculator?
It's about 90 dollars. I never buy those calculators, though. I just get the cheaper versions like TI30x and stuff.
You should look at the inspire or the TI 84 +
Ask a question...how do I use a caculator for this problem
i m from india and here i dont get to use calculator for logs and ln we hv log tables bt i wana knw hw to use 'em!! pls help???
I’m French and at 1:20 ln means in fact "logarithme népérien » !
Will anybody tell how to use log table?
well my log table has log base on top and the logged number on the side. find your base nd your logged number ith your finger and see the number where your fingers meet. that's the answer.
sorry dude i don;t know how to use one?
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