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$$\log_2(x-5)=\log_5(2x+7)$$

Is there a way to solve this equation without drawing the graph and prove they "meet" in one point so the solution is unique? My first idea was to change their base according to the following formula: $$\log_b a = \frac{\log_x a}{\log_x b}$$ but I don't know what base to choose.

Thanks in advance.

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1 Answer

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$$\log_2(x-5)=\log_5(2x+7)=y$$

So $2^y=x-5$ and $5^y=2x+7$. From the first equation, $x=2^y+5$. Substitute this in the second equation to get

$$ 5^y=2^{y+1}+17 $$

We readily see that $y=2$ is a solution.

So $x=2^2+5=9$.

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