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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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  • $\begingroup$ It doesn't matter as long you end up with an equation where you have logs with the same base. Then take anti-logs. $\endgroup$ – herb steinberg Mar 23 at 22:12
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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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  • $\begingroup$ Nice Solution ! $\endgroup$ – Martin Hansen Mar 23 at 22:40

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