[Math] $\log_5 {(x+1)} – \log _4 {(x-2)} = 1$

Question

I tried to solve this equation by changing bases.

$\dfrac {\log_4 (x+1)}{\log_4 5} = \log_4 (4x -8)$

$.86 \log_4 (x+1) = \log_4 (4x-8)$

Then i got stuck.
Please share your idea with me.

Answer 1

As said in comments, I do not think that there is any hope for an analytical solution and numerical methods need to be considered.

Using natural logarithms, the equation write $$\frac{\log (x+1)}{\log (5)}-\frac{\log (x-2)}{\log (4)}=1$$ So, let us consider the function $$f(x)=\frac{\log (x+1)}{\log (5)}-\frac{\log (x-2)}{\log (4)}-1$$ By inspection, $f(3)=\frac{\log (4)}{\log (5)}-1\approx -0.138647$ so the solution is close to $x=3$ (the function being tending to infinity when $x\to 2$).

Probably the simplest method would be Newton starting with $x_0=3$. This would generates the following iterates $$\left( \begin{array}{cc} n & x_n \\ 1 & 2.75504680789501165 \\ 2 & 2.78640480505808886 \\ 3 & 2.78717999430163650 \\ 4 & 2.78718044253601188 \\ 5 & 2.78718044253616154 \end{array} \right)$$