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Find constants such that given equations involving the logarithm hold

  1. Find all constants c such that

        \[ \log x = c + \int_e^x \frac{1}{t} \, dt \]

    for all x > 0.

  2. 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.


  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*}

  2. 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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