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Natural logarithm with a calculator

Natural Logarithm with a Calculator

Natural logarithm with a calculator

Discussion and questions for this video
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.
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."
Where and how does "e" appear in nature?
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?
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).
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)
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 logarithm's argument. You would write that squared logarithm as `ln^2 (2e)`.
Ln actually mean "Logarithm Népérien" reffering to John Napier, who discovered Logarithms. http://en.wikipedia.org/wiki/John_Napier
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))
can someone help me with this equation please
2ln x + ln 5 = 7.2
How do you make the radical or square root sign on your computer?
can anyone help me with this equation
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.
What is an mathematical explation of what e stands for?
e^x is f(x) when (f´(x))/(f(x)=1
not sure if that was that helpful
What I'm more interested in is how to calculate this WITHOUT a calculator
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.
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.
Can someone prove this for me?

lim (a^x -1)/x = ln(a)

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:
(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:
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 cube-root both sides: (5exp[y] + 40)^(1/3) = x, and you've solved for x.
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.
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,
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.
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?
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.
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 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).
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..
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.
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`.
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.
e^0.06t/(1+(0.06/365)^365t = 1000/900
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!
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?
What if e is raised to a power like e-e^(-2x)=1?
how would you find Ln 3x=2? thank and please show me steps to do it.
how do we simplify expressions such as 2^2x4^3x/16x?
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?)
if you use a calculator, it says that sqrt69=8.306623863
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.
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
can u solve this without a calculator?
if your a super genius and can solve decimal problem with only using your mind
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 :)
how do you graph log base 2 of (x + 4) on the ti-83 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.
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
Is there any video that addresses solving or simplifying natural logarithm problems WITHOUT a calculator?
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.
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 solve

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
How do you do natural logs without calculators? Where can I find videos to help me with college-level calculus?
ln is latin for logarithm naturalus
where and how does e appear in nature
How do you Simplify ln(2e)^2 + ln((4e^6)^0.5) - ln(8e^7) without the use of a calculator?
e doesn't show up in nature. In calculus it is the iπ root of -1 because e^iπ=-1 so log -1=iπ
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 TI-30x and stuff.
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 » !
Ask a question...how do I use a caculator for this problem
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.
Report a mistake in the video

At 2:33, Sal said "single bonds" but meant "covalent bonds."

Report a mistake in the video

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