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.
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... and now the quadratic equation... – David G. Stork Oct 15 '19 at 04:47
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Correct, but you have to watch out for extraneous roots. Those would be cases in which $x(5x-2)$ is positive but $x$ and $5x-2$ are negative. You can't take the logarithm of a negative number. (WARNING: DO NOT learn a rule that says a logarithm cannot be negative. That is false. The thing that can't be negative is the number whose logarithm you take.) – Michael Hardy Oct 15 '19 at 04:53
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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.
YiFan Tey
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Yes, sir. I got 2 solutions. They are $\left(\sqrt(406)+1\over5\right)$ and $-\left(\sqrt(406)-1\over5\right)$ which one i should remove ? – bambang nugroho Oct 15 '19 at 04:53
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@bambangnugroho Yes, your thinking is correct. (Misread your comment just now, sorry.) – YiFan Tey Oct 15 '19 at 04:55
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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 $2/5$.
Z Ahmed
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