Why does taking the logarithm of both sides not work?

Hypercubes

The problem statement, all variables and given/known data
Differentiate $x=e^{-x}$

The attempt at a solution
$$\ln{x}=\ln{e^{-x}}$$
$$\ln{x}=-x$$
$$\frac{d}{dx}\left(\ln{x}\right)=\frac{d}{dx}\left(-x\right)$$
$$\frac{1}{x}=-1$$

The correct answer is $1+e^{-x}$. I know how to solve it that way; however, why is the above method wrong? I know there are usually two variables, but I can't see any mathematical errors in my method.

Thank you.

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tiny-tim

Homework Helper
Hi Hypercubes!
Differentiate $x=e^{-x}$

The correct answer is $$1+e^{-x}$$.
Obviously a misprint, for "Differentiate $x-e^{-x}$"

Hypercubes

Well, the exact question was "explain why the equation $e^{-x} = x$ has exactly one solution", and the answer goes on to define $f(x) = x-e^{-x}$ and says "Since $f'(x) = 1 + e^{-x}$, and $1 + e^{-x} > 0$ for all x, the function f is increasing..."

So they set it to 0 to get the derivative. However, why was the method I used not valid?

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D H

Staff Emeritus
Your expression, $x=e^{-x}$ is just that -- an expression. You cannot treat it as a function. In particular, you can't differentiate both sides and expect something meaningful to pop out. What you can do is to rewrite it as $x-e^{-x} = 0$, and now the quantity on the right hand side $f(x)=x-e^{-x}$ can be treated as a function. You aren't differentiating both sides of $f(x)=0$. Just one.

Hypercubes

Ah, I see. Thanks you both!

Matterwave

Gold Member
As a simple example. Take the equation 3x-2=x. This equation has 1 solution x=1. If you differentiate both sides though, you get 3=1 which is obviously not true.

This is because the undifferentiated equation is asking a fundamentally different question than the differentiated equation. The equation 3x-2=x is asking "When is the function f(x)=3x-2 equal to the function g(x)=x?" While the differentiated equation is asking "When is the slope of the function f(x) equal to the slope of the function g(x)?"

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