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I got the following equation to solve for $x$:

$\log_3{(x-3)}= \frac{1}{ 2 \log_2{3} } + \log_{81}(3x-13)^2$

No idea if this is needed but I first tried to bring them to the same base:

$\frac{\log(x-3)}{\log 3} = \frac{\log 2}{2 \log 3} + \frac{\log ((3x-13)^2)}{\log 81}$

Anyone an idea how to solve this? Thanks in advance!

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Multiplying the both sides of $$\frac{\log(x-3)}{\log 3} = \frac{\log 2}{2 \log 3} + \frac{\log ((3x-13)^2)}{\log 81}$$ by $4\log 3\ (=\log 81)$ gives $$4\log (x-3)=2\log 2+\log ((3x-13)^2)$$ So, $$(x-3)^4=2^2(3x-13)^2$$ Use here that $$A^2=B^2\iff (A+B)(A-B)=0$$ with $x-3\gt 0$ and $3x-13\not=0$.

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