Find constants such that given equations involving the logarithm hold

Find all constants $c$ such that
$\log x = c + \int_e^x \frac{1}{t} \, dt$

for all $x > 0$ .
Define
$f(x) = \log \frac{1+x}{1-x}$

for $x > 0$ . Find all $x$ such that

$f(x) = f(a) + f(b)$

where $a$ and $b$ are given constants with $ab \neq -1$ .

We evaluate the integral to solve the equation for the constant,
$\begin{align*} \log x = c + \int_e^x \frac{1}{t} \, dt && \implies && \log x &= c + (\log x - \log e) \\ && \implies && \log x &= c + \log x - 1 \\ && \implies && c &= 1. \end{align*}$
We solve the given equation to find $x$ in terms of $a$ and $b$ ,
$\begin{align*} f(x) = f(a) + f(b) && \implies && \log \frac{1+x}{1-x} &= \log \frac{1+a}{1-a} + \log \frac{1+b}{1-b} \\ && \implies && \log \frac{1+x}{1-x} &= \log \frac{ (1+a)(1+b)}{(1-a)(1-b)} \\ && \implies && \frac{1+x}{1-x} &= \frac{1+a+b+ab}{1-a-b+ab} \\ && \implies && x(2+2ab) &= 2a + 2b \\ && \implies && x &= \frac{a+b}{1+ab}. \end{align*}$

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