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$$\frac{1}{\log_{2x-1}{(x)}} + \frac {1}{\log_{x+6}{(x)}}=1+\frac{1}{\log_{x+10}{(x)}}$$ 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.

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    $\begingroup$ Use $\log_a b=\frac{1}{\log_b a}$ $\endgroup$ – user376343 Oct 25 '19 at 19:26
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Remember - $$\log_{a}b = \frac{1}{\log_ba}$$

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

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

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