Need help to solving the logarithm equation of $\frac{1}{\log_{2x-1}{(x)}} + \frac {1}{\log_{x+6}{(x)}}=1+\frac{1}{\log_{x+10}{(x)}}$

Question

$$\frac{1}{\log_{2x-1}{(x)}} + \frac {1}{\log_{x+6}{(x)}}=1+\frac{1}{\log_{x+10}{(x)}}$$ What should i do for the first step ?

Is it like $\frac{1}{A}+\frac{1}{B}$ then i simplify into $\frac{A+B}{AB}$ ? I need your help or hint to solving this equation. Thank you so much, sir.

Answer 1

Remember - $$\log_{a}b = \frac{1}{\log_ba}$$

Answer 2

Hint: $$\log _ax = {\log x\over \log a}$$

So you have $${\log (2x-1)\over \log x}+{\log (x+6)\over \log x} = 1+{\log (x+10)\over \log x} $$