Find all x satisfying equations given in terms of sinh

Let $c$ be the number such that $\sinh c = \frac{3}{4}$ . Find all $x$ that satisfy the given equations.

$\log (e^x + \sqrt{e^{2x} + 1}) = c$ .
$\log (e^x - \sqrt{e^{2x} - 1}) = c$ .

We are given $\sinh c = \frac{3}{4}$ . From the formula for $\sinh$ this means
$\frac{e^c - e^{-c}}{2} = \frac{3}{4}.$

Then, from the given equation we have

$\log (e^x + \sqrt{e^{2x} + 1}) = c \quad \implies \quad e^x + \sqrt{e^{2x} + 1} = e^c.$

Thus,

$e^{-c} = \frac{1}{e^x + \sqrt{e^{2x} + 1}}.$

So, then we have

$\begin{align*} \frac{3}{4} = \frac{e^c - e^{-c}}{2} &= \frac{1}{2} \left( e^x + \sqrt{e^{2x} + 1} - \frac{1}{e^x + \sqrt{e^{2x}+1}} \right) \\[9pt] &= \frac{1}{2} \left( \frac{(e^x + \sqrt{e^{2x}+1})^2 - 1}{e^x + \sqrt{e^{2x}+1}} \right) \\[9pt] &= \frac{e^{2x} + 2e^x \sqrt{e^{2x}+1} + e^{2x}+1 - 1}{2(e^x + \sqrt{e^{2x}+1})} \\[9pt] &= \frac{e^{2x} + e^x \sqrt{e^{2x}+1}}{e^x + \sqrt{e^{2x}+1}} \\[9pt] &= e^x \end{align*}$

Therefore we have

$e^x = \frac{3}{4} \quad \implies \quad x = \log 3 - \log 4 = \log 3 - 2 \log 2.$
There can be no $x$ which satisfy the given equation. As in part (a), we use the definition of $\sinh x$ to obtain the equation,
$\sinh c = \frac{3}{4} \quad \implies \quad e^c - e^{-c} = \frac{3}{2}.$

Next, we use the equation given in the problem to write,

$\begin{align*} &&\log (e^x - \sqrt{e^{2x} - 1}) &= c \\[9pt] \implies && e^x - \sqrt{e^{2x} -1} &= e^c \\[9pt] \implies && \frac{(e^x - \sqrt{e^{2x} - 1})(e^x + \sqrt{e^{2x}-1})}{e^x + \sqrt{e^{2x}-1}} &= e^c \\[9pt] \implies && \frac{e^{2x} - e^{2x} + 1}{e^x + \sqrt{e^{2x}-1}} &= e^c \\[9pt] \implies && \frac{1}{e^x + \sqrt{e^{2x}-1}} &= e^c. \end{align*}$

Furthermore, we can obtain an expression for $e^{-c}$ by considering

$e^x - \sqrt{e^{2x} - 1} &= e^c \quad \implies \quad \frac{1}{e^x - \sqrt{e^{2x} - 1}} &= e^{-c}.$

Putting these expressions for $e^c$ and $e^{-c}$ into our original equation we have

$\begin{align*} \frac{3}{2} &= e^c - e^{-c} \\[9pt] &= \left( \frac{1}{e^x + \sqrt{e^{2x}-1}} \right) - \left( \frac{1}{e^x - \sqrt{e^{2x}-1}} \right) \\[9pt] &= \frac{e^x - \sqrt{e^{2x}-1} - e^x - \sqrt{e^{2x}-1}}{(e^x + \sqrt{e^{2x}-1})(e^x - \sqrt{e^{2x}-1})} \\[9pt] &= \frac{ -2 \sqrt{e^{2x}-1}}{e^{2x} - e^{2x} + 1} \\[9pt] &= -2\sqrt{e^{2x}-1} \end{align*}$

But this implies

$\sqrt{e^{2x}-1} = -\frac{3}{4}$

which is impossible. Hence, there can be no real $x$ satisfying this equation.