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I have done many $\log$ problems but I've never learned something such as $\log_ax-\log_by$. I know that to condense a logarithm you must have the same base: $\log_ax-\log_ay=\log_a\left(\frac{x}{y}\right)$. With that said,

Simplify the following expression:
$2\log_49-\log_23$ . Which now becomes, after change of base, $\log_481=\dfrac{\log_43}{\log_42}$. Which makes $\log_481=\log_4\left(\frac{3}{2}\right)$ Is this right so far? And what to do after this.

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    You're welcome. You might want to go through the [identities page](http://en.wikipedia.org/wiki/List_of_logarithmic_identities) on wikipedia.2012-07-15

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Hints:

  1. $\log_a x = \frac{\log_b x}{\log_b a}$

  2. $c\ \log_a x = \log_a x^c.$

You can use that to unify the base.

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    Yes. If you want your unified base to be 4.2012-07-15
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Recall that $x=\log_4(9)$ is defined to be the unique number with the property that $4^x=9$.

Because $4=2^2$ and $(a^b)^c=a^{bc}$, we therefore have that $(2^2)^x=2^{2x}=9$, which by definition means that $\log_2(9)=2x$.

Now use that $\log_b(t^2)=2\log_b(t)$ for any $t>0$.

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    That's a correct statement, but you're going the wrong way; you want to extract a $\log_2(3)$ somehow.2012-07-15
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$2*\log_49-\log_23$ Use $\log_a x = \frac{\log_b x}{\log_b a}$, $2*\frac{\log_29}{\log_24}-\log_23$ $2*\frac{\log_29}{2}-\log_23$ $\log_29-\log_23$ $\log_2{\frac{9}{3}}$ $\log_23$