No real solution to logarithmic equation?

Question

$$e^x + 1 = 2e^{-x}$$

Wolfram Alpha claims no real solution and my text book claims the solution $x=0$.

Why can't I simply multiply each side by $e^x$:

$$e^{2x} + e^x = 2$$

$$\ln(e^{2x}) + \ln(e^x)=\ln(2)$$

$$3x = \ln(2)$$

$$x = \frac{\ln(2)}{3}$$

??

Answer 1

$\log(a+b) \neq \log(a)+ \log(b)$.

So, $\log(e^{2x}+e^x) \neq 3x$.

Answer 2

As stated above you cannot do what you have done.

You must put $e^x=t$ and then form a quadratic in $t$. Then, you have to consider positive roots as $e^x>0$.

This will give you $t=-2,1 \implies t=1 \implies e^x=1 \implies x=0$.

And never doubt wolfram alpha