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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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    Just edited it with what I believe is correct.2012-07-15
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    This is not correct. As indicated below: $2\log_4 9 = \log_4 81.$ What did you do to convert $\log_2 3 = \log_4 x$??2012-07-15
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    $\log_a x = \dfrac{\log_b x}{\log_b a}$ with $a=2$ and $b=4$ and $x=3$ correct?2012-07-15
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    Right. That's $\log_2 3 = \frac{\log_4 3}{\log_4 2} = \frac{\log_4 3}{1/2}.$ Can you simplify it properly now?2012-07-15
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    So, you can't have, $\log_23=\frac{\log_43}{\log_42}$ and that can't equal $\log_4\left(\dfrac{3}{2}\right)$???2012-07-15
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    Or does it become $\log_4(3-2)$2012-07-15
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    No. $\frac{\log x}{\log y} \neq \log{\frac{x}{y}}.$ What you should do is simplify the $1/2$ fraction (Remember $\frac{x}{1/2} = 2x$?) and use the fact that $2\log X = \log X^2.$2012-07-15
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    $\log_23=\frac{\log_43}{\log_42}=\frac{\log_43}{1/2}=2\log_43=\log_43^2=\log_49$??2012-07-15
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    Yes. Then $2\log_4 3 = \log_4 3^2.$2012-07-15
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    So, $\log_4 81-\log_49$?2012-07-15
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    Yes. Now you can simplify into a fraction.2012-07-15
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    Final answer: $\log_4\left(\dfrac{81}{9}\right)=\log_4 9$?2012-07-15
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    That's correct!2012-07-15
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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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    $a=2$ and $b=4$ ?2012-07-15
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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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    so that makes $2\log_49=\log_481$ ?2012-07-15
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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 $\log_a x = \frac{\log_b x}{\log_b a}$, $$2*\frac{\log_29}{\log_24}-\log_23$$ $$2*\frac{\log_29}{2}-\log_23$$ $$\log_29-\log_23$$ $$\log_2{\frac{9}{3}}$$ $$\log_23$$