4
$\begingroup$

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.

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Stefan4024, José Carlos Santos, Zhanxiong, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 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
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $x \approx 3.70931751$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.