How to evaluate a natural logarithm using a calculator (example)

Sal evaluates log_e(67) (which is more commonly written as ln(67) ) using a calculator.

How to evaluate a natural logarithm using a calculator (example)

Discussion and questions for this video
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."
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 higher-level math.
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.
When Sal says e shows up in nature a lot, what does he mean? Where in nature does e show? Like, I know pi, in nature, is the ratio of circumference to diameter, is there any such thing for e?
In my work, I encountered e a lot more than π. The constant shows up in exponential functions all the time, such as in radioactive decay. It even shows up in such things as statistics, business math, civil engineering, and computing interest -- just to name a few.

As far as why we would use such a strange number as e as the preferred base for a logarithm, that will become very evident if you go on to study calculus. For now, let us just say that the math is so much easier with the natural logarithm that in higher levels of math and many applied uses of math, the natural log is used almost to the exclusion of any other base. There are some specialized fields where it makes more sense to use a base 2 or base 10 logarithm, but the natural log is far, far easier in the vast majority of applications.
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?
Try dividing both sides by 4 first.
e^x = 10/4
Now can you solve?
ln is the natural logarithm. It is log to the base of e.
e is an irrational and transcendental number the first few digit of which are:
In higher mathematics the natural logarithm is the log that is usually used. The log on your calculator is the common log, which is log base 10.
"e" is the natural representation for any problem involving exponential growth. For example, half-life 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).
If e is truly infinite, how can one tell that it _never_ repeats? You'll never reach the end of the number, and thus, you won't now the number's true value to know if it repeats.
A number that repeats digits in its decimal form is a rational number meaning that it can be represented as the quotient of two whole numbers. There is a high level proof on this link that e is irrational:
Where exactly is the number "e" found in nature? Why do people call it a natural number?
e comes up all the time in real-world math. For example, it is used in business math for certain kinds of interest calculations. It is used in calculating certain kinds of reaction rates, especially radioactive decay. It is used extensively in engineering computations.
Does anyone know how to change the base on a TI-83 to something other than ten?
actually ON ALL CALCULATORS you have to do this equation, log a(b) = log b(b)/log b(a)
What if you have a question that asks like e^2x+1=55? How can you solve for e?

"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).
What is e and why is it used so much, and where? I know it is like pi, both are just numbers that go on forever. Pi is about 3.1415, and it is used with circles. You need pi to find the area and circumference.
E is about 2.71, but where is e used and why is it so commonly used as a base for logarithms?
`e` or `Euler's number` is a very important mathematical constant that is found in many things. It estimates to be about `2.71828`... It is the base of natural logarithms. `e` is used in calculus, complex numbers, chemistry and much more.

You can use a binomial expansion to calculate `e`. For example take a look at `(1 + 1/n)^n` if this `n` approaches infinity or is extremely large it approaches the value of `e`. If n = 100000 then the output is `e` correctly only to the 4th decimal place.

`e` is the limiting factor used in all kinds of scientific processes and economic functions.
Yes it can! e^x = 1 + x + x^2/2 + x^3/3! + x^4/4!..... or you can express it as lim n-->infinity (1+1/n)^n
`e` or `Euler's number` is a very important mathematical constant that is found in many things. It estimates to be about `2.71828`... It is the base of natural logarithms. `e` is used in calculus, complex numbers, chemistry and much more.

You can use a binomial expansion to calculate `e`. For example take a look at `(1 + 1/n)^n` if this `n` approaches infinity or is extremely large it approaches the value of `e`. If n = 100000 then the output is `e` correctly only to the 4th decimal place.

`e` is the limiting factor used in all kinds of scientific processes and economic functions.
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.
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.
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))
ACTUALLY ↑←hkapur97→↓ e^iπ =-1 and e=lim_x→∞(1+1/x)^x
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*[(x-1)/(x+1))]^(2n-1)/(2n-1)

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.
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.
Can someone give an explanation of e? What is its significance? Who discovered it? (etc.) Or, if there is a video I did not happen to notice on this subject, can someone direct me to a tutorial? Thanks!
Euler's number, commonly known as 'e', is no less important than Pi or any other mathematical constant. The only thing is that we don't bump into it as often, at least while we're still in high school, or pursuing a career in a completely different field.

The discovery of the constant itself is credited to Jacob Bernoulli, who attempted to find the value of the following expression (which is in fact e): (

Its significance? There are many. Some of them are:

1) "e" is the base of Natural Logarithm (ln). Given that x=ln (y), then y = e^x
2) "e" is also used in complex numbers, e^(ix) = cos x + i sin x.
3) "e" has a special place in Calculus, d/dx (e^x) = e^x, d/dx (ln x) = 1/x
4) "e" is also used in the definitions of hyperbolic functions, sinh x , cosh x and tanh x
5) "e" is the limit of (1 + 1/n)^n as n approaches infinity, an expression that arises in the study of compound interest.

You may want to check out this Wikipedia article for more info:

Hope this helps! :)
e^x-5*e^(-x)=4-->e^x-5*(1/(e^x))=4-->e^x-5/(e^x)=4. Let e^x be n.
n-5/n=4 (Multiply each side by n)-->n^2+5=4n (Subtract each side by 4n)-->n^2-4n+5=0 (Factor)-->(n-5)(n+1)=0.
n-5=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.
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 cube-root both sides: (5exp[y] + 40)^(1/3) = x, and you've solved for x.
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,
How would I go about finding the number whose natural log is a given number? e.g. 2.76
So really my question is, how would I work out 'x' of: ln x = 2.76
This is not covered in this section that I can see?
You can figure it out just by trial and error trying numbers in but that might take a long time. And what if the answer is in decimals? You'll probably have a better chance of getting the answer faster with a calculator. Also, a better chance to get it correct.
Unless you are very advanced in your math skills, the short answer is that you don't.
The standard custom at this level of study is either express your answer in terms of the natural log, to use a calculator or other computing device, or you will be given the logs you need.

There is a way to solve natural logs by hand, of course, but it is quite difficult math and you will not be asked to do it at this level of study. The only exception would be that you might be expected to recognize those special cases where the natural log has a nice, easy solution (such the natural log of e raised to some exponent).

But, just for reference, the actual computation is:
ln x = lim h→0 [x^(h) − 1] / h
There are other ways to compute the natural logarithm by hand, none of them easy.
At 2:55, I cannot understand how does ln(67)=4.205 approximately matches between 2 and 3. plz help:(
If we let ln67 = a
then e^a = 67 by definition of a logarithm
e by definition is about 2.71 I think
If we round the 4.2 down for approx.
2^4 = 16 and 3^4 = 81
Therefore since e = 2.71 which is in between 2 and 3
we expect e^4 to be in between 16 and 81, which it is.
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:
(a⁰-1) / 0 = (1-1) / 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:
Could someone please tell me where can I obtain the same emulated graphing calculator that Sal is using? (Please include the download link too!)
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.
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?)
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?
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 :)
One thing that I find hard to understand ( altought it may seem a bit silly ) is how can you have a number raise to a non-integer number.... I mean, when you like 2², you multipply 2 times 2. How can you multiply, like, 4.3012312371 times 2? It isn´t really intuitive
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 in-depth on how the e^ln or ln e = 1 works?
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 same-base exponent and logarithm:
5x = e⁴
x = ⅕e⁴
Thus x is an irrational and transcendental number. It is approximately:
x ≅ 10.919630
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..
at 2:00 he "pulls" out his calculator which is an emulator for the ti-85 i guess. does anyone know how to get the software? for mac would be cool.
Im not understanding the meaning of "e". Where is the derivation or proof shown for e?How come e is 2.71....?
e is a constant that is defined by limits, so if you have not yet studied limits, the definition of e won't mean much to you. Since e is irrational and transcendental, it is impossible to express it exactly by algebraic operations on rational, real numbers.

Here is the definition of e:
e = lim h → 0 (1+h)^(1/h)
how is log different than e?
When and why do you use e?
If you are given an equation without e how are you supposed to use e to love it?
Ex: 5^2x=38
The natural log function is the inverse of e. Much like dividing is the inverse of multiplying, so if we had something in terms of e we could use the natural log to solve it. The natural log is the log of base e, so in the example you gave we wouldn't need it since the base isn't e.
What does log 3 x mean? Is it like base 3 with power x or you just multiply 3 with x?
Your question is very ambiguous.

`log_3(x) = y` is equivalent to `3^y = x`.

If you were solving for `y` then:

3^y = x

ln(3^y) = ln(x)

y•ln(3) = ln(x)

y = [ln(x)/ln(3)]

If your expression was instead `log(3^x)` then this simpifies to:

What is e, as a concept? I know it is used for logs and it equals 2.71... but what is the reason for it?
What is the exact calculator Sal is using? I know it's a Texas Instruments calculator but I don't know which type it is.
I am confused when I try to extrapolate this to create an equivalent exponential equation from a logarithmic equation, such as e^2=x+1. Which video should I go to to explain this? Thank you.
ln2.logb5^4 = log1016.ln10
is it possible to club ln and log or we have to treat them differentely.
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.
I do not know what 'e' really is. Is there a video by Sal where he tells what e actually is? In this video he is only telling us how to solve for log_e!
Dont be bothered with the 'e' its a number lyk 'pi'
And its value is some 2.718281828459045 ....already mentioned by just...
If you use the Change-of-Base formula for log base e of 67:
log base e of 67 = log (67) / log (e)
would you get the exact same answer to ln (67)?
How would you solve log base 5 (13) by using common logs or natural logs and a calculator? Would I use e somehow? -1:06 minutes
you can use either the log (base 10) button or the ln (base e) button. I prefer the natural log for no particular reason.

log_5 (13) = ln 13 / ln 5 = 1.59...

This is a result of the "Change of Base" formula.
a logarithm is essentially a different way of writing an exponent. It is written like this:

log a (b)=x

a real example of this is:

log 3 (9)=2, because
each log has a base (in this case, a). in an "ln" or natural logarithm, the base is e. so ln essentially means:
e raised to the xth power makes the number we want.
From where can we get the TI-85 which Khan Academy uses. If an emulator is there please tell me where i can download it from.
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.
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).
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.
How do you do natural logs without calculators? Where can I find videos to help me with college-level calculus?
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 you Simplify ln(2e)^2 + ln((4e^6)^0.5) - ln(8e^7) without the use of a calculator?
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