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Find values of x satisfying equations with logarithms

Find an x \in \mathbb{R} such that the following equations hold.

  1. \log (1+x) = \log(1-x).
  2. \log (1+x) = 1 + \log(1-x).
  3. 2 \log x = x \log 2 for x \neq 2.
  4. \log \left( \sqrt{x} + \sqrt{x+1} \right) = 1.

  1. We compute,

        \begin{align*}  \log(1+x) = \log(1-x) && \implies && 1+x &= 1-x \\  && \implies && 2x &= 0 \\  && \implies && x &= 0. \end{align*}

  2. We compute,

        \begin{align*}  \log(1+x) = 1+\log(1-x) && \implies && \log \frac{1+x}{1-x} &= 1 \\  && \implies && \frac{1+x}{1-x} &= e \\  && \implies && x & = \frac{e-1}{e+1}.  \end{align*}

  3. We compute, for x \neq 2,

        \begin{align*}  2 \log x = x \log 2 && \implies && \log (x^2) &= \log (2^x) \\  && \implies && x^2 &= 2^x \\  && \implies && x &= 4. \end{align*}

  4. We compute,

        \begin{align*}  \log \left( \sqrt{x} + \sqrt{x+1} \right) = 1 && \implies && \sqrt{x} + \sqrt{x+1} &= e \\  && \implies && x - (x+1) &= e \left( \sqrt{x} - \sqrt{x+1} \right) \\  && \implies && \sqrt{x+1} - \sqrt{x} &= \frac{1}{e} \\  && \implies && \left( \sqrt{x+1} - \sqrt{x} \right) + \left( \sqrt{x} + \sqrt{x+1} \right) &= \frac{1}{e} + e \\  && \implies && 2 \sqrt{x+1} &= e + \frac{1}{e} \\  && \implies && \sqrt{x+1} &= \frac{e}{2} + \frac{1}{2e} \\  && \implies && x+1 &= \frac{e^2}{4} + \frac{1}{4e^2} + \frac{1}{2} \\  && \implies && x &= \frac{e^2}{4} + \frac{1}{4e^2} - \frac{1}{2} \\  && \implies && x &= \frac{e^4 + 1 - 2e^2}{4e^2} \\  && \implies && x &= \frac{(e^2-1)^2}{4e^2}. \end{align*}

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