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Solve for x;

$\log_{12}x=\frac{1}{2}\log_{12}9+\frac{1}{3}\log_{12}27$

The only thing throwing me off is the one third and one half, which my book does not say how to fix.

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  • $\begingroup$ Hint: $\log x^y=y\log x$. $\endgroup$ – vadim123 May 2 '13 at 3:14
  • $\begingroup$ You can use the algebraic properties of logarithms to simplify the right side. $\endgroup$ – Jonas Meyer May 2 '13 at 3:14
  • $\begingroup$ and $\log_c a+\log_c b=\log_c ab$ $\endgroup$ – Ehsan M. Kermani May 2 '13 at 3:16
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$$\log_{12}9=\log_{12}3^2=2\log_{12}3$$

Similarly, $$\log_{12}27=\log_{12}3^3=3\log_{12}3$$

So, $$\log_{12}x=2\log_{12}3=\log_{12}3^2=\log_{12}9\implies x=9$$

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$\log_{12}x=\log_{12}9^{0.5}+\log_{12}27^{0.3333....}\implies\log_{12}3+\log_{12}3\implies \log_{12}9= \log_{12}x$. Therefore, $x=9$. Sorry for my horrible typing, I don't know how to use that program to make the writing nicer.

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