Verifying $\ln(1 + e^{2x}) = 2x + \ln(1 + e^{-2x})$

Question

The book is simply asking me to show that this equation is true:

$$\ln(1 + e^{2x}) = 2x + \ln(1 + e^{-2x})$$

Answer 1

$$\ln(1 + e^{2x}) = 2x + \ln(1 + e^{-2x})$$ Exponentiate both sides, i.e., raise $e$ to these powers on either side $$(1 + e^{2x}) = e^{2x} \cdot (1 + e^{-2x})=(e^{2x}+1) $$ okay

Answer 2

Hint: note that $$1+e^{-2x} = \frac{e^{2x}+1}{e^{2x}}$$ and use $\ln\bigg{(}\frac{a}{b}\bigg{)} = \ln a - \ln b$.

Answer 3

$$ 1 + e^{2x} = e^{2x} \left( 1 + e^{-2x}\right) $$ Which is true.

Truth is just truth. You can't have opinions about truth.

The first two and a half minutes of https://www.youtube.com/watch?v=4MOAtq91idk

Answer 4

Write

$\ln(1 + e^{2x}) = 2x + \ln(1 + e^{-2x})\\ \Leftrightarrow \ln(1 + e^{2x}) - \ln(1 + e^{-2x})= 2x \\ \Leftrightarrow \ln(\frac{1 + e^{2x}}{1 + e^{-2x}})= 2x $

and exponentiate on both sides to obtain

$\frac{1 + e^{2x}}{1 + e^{-2x}}=e^{2x}\\ \Leftrightarrow 1 + e^{2x}=(1 + e^{-2x})e^{2x}=1 + e^{2x}$

which is true for all x in $\mathbb{R}$ or even $\mathbb{C}$

Answer 5

Hints: $\quad \ln a + \ln b = \ln(a+b), \quad \ln a^n = n\ln a, \quad \ln e = 1, \quad a^na^m=a^{n+m}, \quad a^{0}=1, a \neq 0$

So, we can write: $$2x = \ln e^{2x}$$

Thus, we have: $$\ln(1+e^{2x})= 2x + \ln(1+e^{-2x}) \\$$ $$\iff \ln(1+e^{2x})= \ln e^{2x} + \ln(1+e^{-2x})$$ $$\iff \ln(1+e^{2x})= \ln e^{2x}(1+e^{-2x})$$ $$\iff \ln(1+e^{2x})= \ln (e^{2x}+e^{0})$$ $$\iff \ln(1+e^{2x})= \ln (e^{2x}+1)$$