How can I find $x$ from this logarithms equation? Solve $\log_9x + \log_{81}3x = 1$
$\log_9x=2\log_{81}x$

So, $$81^{2\log_{81}x+\log_{81}3x}=81$$$$x^2\cdot3x=81$$$$x^3=27$$

So, if $x\in\mathbb R, x=3$
Hint:  $\log_9 x= \log_{81}x^2$