gradient of natural log

[enginist]

I can't satisfy myself that the gradient of logX for base e at the argument X is equal to 1/X


As I understand it, the gradient or slope of any tangent to a curve is the vertical rise over the horizontal distance, or Y/X


But when I try to work it out for the function logX, I can't get it right


For example, the gradient of logX for X=10 should be 1/10


But if I use the Y/X method to calculate the slope, I get X=10 and Y=2.302, since the log of 10 for the base e is 2.302


Plugging this into Y/X, I get 2.302/10, which clearly does not equal 1/10


I know I'm doing something wrong, but after several hours of effort I don't know what it is

 

[pka]

I can't satisfy myself that the gradient of logX for base e at the argument X is equal to 1/X

Why do you think that?
By definition if \(\displaystyle x>0\) then \(\displaystyle \displaystyle\log (x) = \int_1^x {\frac{{dt}}{t}} \).
So if you understand how things are properly defined there is no doubt that \(\displaystyle D_x(\log(x))=\dfrac{1}{x}~?\)

Now what about that do you not understand?

 

[soroban]

Hello, enginist!

I can't satisfy myself that the gradient of log X for base e at the argument X is equal to 1/X.

Who told you that?
If you are in Intermediate Algebra, you can't possibly know that!

First, you have to be familiar with natural logs (logs to the base e), written \(\displaystyle \ln x.\)

Then, in Calculus, you would learn that the derivative of \(\displaystyle \ln x\) is \(\displaystyle \dfrac{1}{x}\)

. . and that the derivative gives us the slope of the tangent (gradient) to the graph.

 

[enginist]

Thanks for responding, PKA (sounds like a fraternity).

Because I'm not there yet, let's forget about derivatives for a while and just consider this in terms of a simple gradient. The gradient of any tangent to a curve is horizontal over vertical, or Y/X, correct?

If I pick X=10 for the function logX, then Y=2.302, since log10 for base e is 2.302.

So the gradient of the tangent at the point where X=10 and Y=2.302 is Y/X, or 2.302/10, which clearly doesn't equal 1/10.

The error is either in my thinking or in the present state of my knowledge, which admittedly is shallow. Can I find a gradient of logX without knowing calculus? Can I determine a slope of the curve of e without applying derivatives?

--Tim

I and the public know
What all school children learn,
Those to whom evil is done
Do evil in return.

--W.H. Auden

 

[enginist]

Thanks for responding, soroban.

You're right, I wouldn't know that if I were still a sophomore in high school. But I am a retired fireman trying to learn a little math. Specifically, I'm studying a book for the curious layman about the Riemann Hypothesis. The book is very thorough and I am making steady progress, although I often need to stop in order to fill the gaps in my knowledge. This can make it hard sometimes, but at least I have the luxury of time. I don't know calculus and what I know about algebra I'm learning on the fly. And that's why I came here.

--Tim

 

[enginist]

Yes, it helps. The error was in both my thinking and my knowledge.

Thanks, Jeff.

--Yim