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I have a logarithmic equation that I am meant to find the value of $x$ for:

$2\log_{9}(x)$ = $1/2$ + $2\log_{9}(5x+18)$

I get as far as here when I realize I cannot use the quadratic formula.

$\log_{9}\left( \dfrac{x^{2}}{(5x+18)^2}\right)= \dfrac{1}{2}$

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But you know what to do with $\log_9(x^2)=\log_9(3)+\log_9((5x+18)^2)$, yes? – J. M. Nov 9 '11 at 11:10

From (note, I corrected a slight error) $$ \log_9\Bigl( ({x\over 5x+18})^2\Bigr)={1\over2}. $$ Get rid of the logarithm, by writing $$ ({x\over 5x+18})^2 =9^{1/2}, $$ and proceed from there.

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I should point out that what I used in the above was $\log_a x=y\iff x=a^y$. – David Mitra Nov 9 '11 at 11:20

No need to introduce squares. Simplify the original equation to $$ \log_9({x\over 5x+18})={1\over4} $$ and then to $$ {x\over 5x+18}=9^{1\over4}. $$

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