# logarithm proof fallacious or not?

$e^{-x}=e^{1/x}$ Taking the natural logarithm of both sides

$$\ln(e^{-x})=\ln(e^{1/x})$$

$$-x=1/x$$

$$-x^2=1$$

$$x^2=-1$$

$$x=i$$ I know I am doing something wrong here. Also can someone please explain why $$-\ln(x)=\ln(1/x)$$ Thank you.

• As already answered, your premise $e^{-x}=e^{1/x}$ is incorrect. On the other hand, if you think of it as an equation, it seems that $x=i$ would be a correct solution. – Mirko Jul 28 '15 at 22:53
• If $x^2 = -1$, then $x=\pm i$, but the fact that $x^2=-1$ is not enough to tell you which of those two it is. ${}\qquad{}$ – Michael Hardy Jul 28 '15 at 23:14

First statement is wrong. $$e^{-x} = \frac{1}{e^x} \neq e^{1/x}$$

As for the other question, note that if you exponentiate both sides, you get $$e^{-\ln x} = \frac{1}{e^{\ln x}} = \frac{1}{x}$$ on the left and the same thing on the left. Certainly in real numbers, $x=y$ iff $e^x=e^y$.

It's true that $e^{-x}=1/(e^x)$, but $e^{-x}=e^{1/x}$ is false, hence the 'proof' is meaningless.

Where you wrote $e^{-x} = e^{1/x}$ I thought at first that you meant you were trying to solve that equation for $x$. If one could assume $x$ is real, then that would entail that $x^2=-1$ so that $x=\pm i$. But that would mean $x$ is not real. If we allow $a$ and $b$ to be complex then $e^a = e^b$ no longer entials that $a=b$, but only that $a = b+n\pi i$ for some integer $n$ Then you would have $$-x = \frac 1 x + n\pi i,$$ so $$x^2 = -1-n\pi i$$ and you'd go on from there.

However, it seems you may have meant that for all values of $x$ it is the case that $e^{-x}= e^{1/x}$. That is incorrect. In fact $$e^{-x} = \frac 1 {e^x}$$ but that's not at all the same thing as $e^{1/x}$.

Notice, your first step $e^{-x}=e^{1/x}$ is wrong. The correct one is $$e^{-x}=\frac{1}{e^{x}}$$ Then you may take log as $$\ln e^{-x}=\ln\left(\frac{1}{e^x}\right)$$

I assume your are trying to solve $e^{-x}=e^{1/x}$ in the unknown $x$. So using only properties of the exponential, you get $e^{x+1/x} = 1 = e^0$. You are no longer in the standard real case, since the sum of two reals with the same sign ($x$ and $1/x$) cannot be 0. This shows there is a risk in taking (unnecessarily) the logarithm of an expression that could be something else than a positive number. This is a manifestation of the economy principle in proving in mathematics. You have to change the field...

Standard complex solutions to $x+1/x = 0$ are$x_1 = \imath$ and $x_2 = -\imath$. But there is more to that. First, as pointed out by Michael Hardy, $e^{\imath k \pi} = (-1)^k$ for any integer $k$, and you can get more solutions. Second, you can got beyond the complex field, and look for solutions in other domains, for instance quaternions, see e.g. Exponential Function of Quaternion - Derivation, octonions and less standard structures in Musean hypernumbers.

As for the second part of the question: two key properties of the natural logarithm (with positive and real numbers) are:

1. $\ln(1) = 0$,
2. $\ln(ab) = \ln(a) + \ln(b)$.

Hence, for any positive $x$, $0=\ln(1) = \ln(\frac{x}{x}) = \ln(x)+\ln(\frac{1}{x})$. Rejoining the two ends of the equations yields your answer.