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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*}

4 comments

    • Artem says:

      I am also interested if it can be solved analytically.
      But I think here the method of “informed guess” was taken – we just guessed the number. I personally checked the graph plotter to see this.

      • TRC YX says:

        I searched on WolframAlpha, and the solution is analytically represented by a lambert W function, where W is the inversion of f(x) = xe^x, so that

            \[W(x)e^{W(x)} = x,\]

        and it’s impossible to express this W in terms of elementary functions.

        The equation 2^x=x^2 can be separated into two equations \sqrt{2}^x=x and \sqrt{2}^x=-x. Since \log is only defined for positive values, we can only solve the first one (while the second one does yield a negative real solution).

            \[<span class="ql-right-eqno"> (1) </span><span class="ql-left-eqno">   </span><img src="http://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ae8977549e6c0b2ab364508991b19a70_l3.png" height="272" width="447" class="ql-img-displayed-equation quicklatex-auto-format" alt="\begin{align*} &\sqrt{2}^x &= x \\ \Longrightarrow &1 &= \frac{x}{\sqrt{2}^x} \\ \Longrightarrow &1 &= x\sqrt{2}^{-x} \\ \Longrightarrow &1 &= x\exp\left(-x \cdot \frac{\log(2)}{2}\right) \\ \Longrightarrow &-\frac{\log(2)}{2} &= \left(-x \cdot \frac{\log(2)}{2}\right) exp\left(-x \cdot \frac{\log(2)}{2}\right) \\ \Longrightarrow & W\left(-\frac{\log(2)}{2}\right) &= -x \cdot \frac{\log(2)}{2} \\ \Longrightarrow & -\frac{2W\left(-\frac{\log(2)}{2}\right)}{\log(2)} = x \end{align*}" title="Rendered by QuickLaTeX.com"/>\]

        Since multiple x can have the same xe^x, so there are two branches of W, and hence two solutions of the above equation, which are 2 and 4 respectively, and 4 is the answer we need here.

        However I’m not sure how to look through the expression of the solution to see 2 and 4, other than guessing the value of W. (

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