# Different log bases

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. – Austin Broussard Jul 15 '12 at 20:24
This is not correct. As indicated below: $2\log_4 9 = \log_4 81.$ What did you do to convert $\log_2 3 = \log_4 x$?? – user2468 Jul 15 '12 at 20:27
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? – user2468 Jul 15 '12 at 20:29
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 $2\log X = \log X^2.$ – user2468 Jul 15 '12 at 20:34
Yes. Then $2\log_4 3 = \log_4 3^2.$ – user2468 Jul 15 '12 at 20:37

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 $b=4$ ? – Austin Broussard Jul 15 '12 at 20:11
Yes. If you want your unified base to be 4. – user2468 Jul 15 '12 at 20:13

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 $2\log_49=\log_481$ ? – Austin Broussard Jul 15 '12 at 20:12
That's a correct statement, but you're going the wrong way; you want to extract a $\log_2(3)$ somehow. – Zev Chonoles Jul 15 '12 at 20:12

$$2*\log_49-\log_23$$ Use $\log_a x = \frac{\log_b x}{\log_b a}$, $$2*\frac{\log_29}{\log_24}-\log_23$$ $$2*\frac{\log_29}{2}-\log_23$$ $$\log_29-\log_23$$ $$\log_2{\frac{9}{3}}$$ $$\log_23$$

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I downvoted. The question is tagged (homework). We don't just give away answers to homeworks. We instruct OP through the answers. – user2468 Jul 15 '12 at 22:41
@J.D.: Sorry for that. This mistake will not be repeat next time. But I think this shouldn't be reason to downvoted. – rekenerd Jul 15 '12 at 22:48
Sorry for being harsh. I revoked my downvote. – user2468 Jul 15 '12 at 22:52
@J.D.: Thanks and I will remember this point to give just hints in homework tagged questions. – rekenerd Jul 15 '12 at 22:56