How can I solve it?
$$ \frac{2}{\log_{8}(x-1)} - \frac{2}{\log_{8}(x )} =1$$

I don't have idea how to solve it and I will be happy for help about this exercise.


closed as off-topic by Namaste, Stefan4024, José Carlos Santos, Zhanxiong, Parcly Taxel Dec 6 '17 at 3:11

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  • 1
    $\begingroup$ Multiply the fractions crosswise, and rewrite $1$ as a log $\endgroup$ – R3E1W4 Dec 5 '17 at 16:06
  • 3
    $\begingroup$ @R3E1W4: I fail to see how this helps... $\endgroup$ – Yves Daoust Dec 5 '17 at 16:21
  • $\begingroup$ Where do this equation come from? $\endgroup$ – Raffaele Dec 5 '17 at 16:37

This problem does not have a closed-form solution.

Combine the fractions: $$ \frac{2\log_8 x - 2\log_8 (x-1)}{\log_8 x \log_8 (x-1)} =1 \\ \log_8 x - \log_8 (x-1) = \frac12 \log_8 x \log_8 (x-1) $$ Then use properties of the logarithm: $$ \log_8\left( \frac{x}{x-1} \right) = \log_8 \sqrt{ (x-1)^{\log_8 x }} $$ Now raise $8$ to both sides of this equation and square both sides: $$ \frac{x^2}{(x-1)^2} = (x-1)^{\log_8 x}$$ Handle the denominator on the left by adding to the exponent on the right: $$x^2 = (x-1)^{2+\log_8 x}$$

This has a unique real positive solution at roughly $x = 3.7093175$, but there is no solution in closed form using only elementary functions.


One can shift the equation by using: \begin{align} \log_{b}(x) &= \frac{\log_{d}(x)}{\log_{d}(b)} \\ x &\to t + \frac{1}{2} \end{align} to obtain $$\frac{1}{\ln\left(t - \frac{1}{2}\right)} - \frac{1}{\ln\left(t + \frac{1}{2}\right)} = \frac{1}{6 \, \ln(2)}.$$ The solution for $t$ is $t \approx 3.20931751$ and yields $x \approx 3.70931751$.


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