Solving logarithm equation of $\log_{3}(5x-2)+\log_{3}(x)=4$

Question

My work so far is $$\log_{3}(5x-2)+\log_{3}(x)=4$$ $$\log_{3}(5x-2)+\log_{3}(x)=\log_{3}(81)$$ $$\log_{3}\left(x(5x-2)\right)=\log_{3}(81)$$ $$5x^2-2x-81=0$$

Is it correct so far ? Thanks for your help and suggestion.

Answer 1

Yes, it's correct, except that you need to ensure $x>0$ and $5x-2>0$. This is implicit in the question, since the logarithm only takes (strictly) positive arguments. So solving the quadratic equation you got, you will end up with two answers, but one of them will be negative and hence needs to be rejected.

Answer 2

First declare that $x>2/5$, then $$\log_{~3} (5x-2) +\log_{~3} x=4 \implies \log_{~3} [x(5x-2)]=\log_{~3} 3^4\implies 5x^2-2x-81=0 \implies x=\frac{2\pm \sqrt{1624}}{10}$$ So only one root as $x=\frac{2+\sqrt{1624}}{10}\approx 4.22$ is positive and greater than $2/5$.