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Write as a single logarithm: $\log_8(5) - 2\log_8(6)$

To my understanding; because they are the same base you can just evaluate $\log_8\left(\frac{\log(5)}{\log(6)}\right)$ which is shown on the multiple choice I have as $\log_8\left(\frac{5}{12}\right)$.

However this is apparently not the correct answer. Where did I go wrong?


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I've added LaTeX formatting to your question; apologies if I changed your intended meaning in any way. – Zev Chonoles Aug 20 '11 at 21:11
Thanks so much. Can you tell me where I can inform myself upon how to format the texts like that? – John Aug 20 '11 at 21:13 This is a good reference if you want to get familiar with $\LaTeX$ in general. If you just want to get familiar with latex for this site, then I suggest you just search `Math in latex' on google and you will probably get some references. – sxd Aug 20 '11 at 21:17
You can also right click on a piece of $LaTeX$ in a question, answer, or comment to see the $LaTeX$ code that goes between the \$'s. – robjohn Aug 20 '11 at 21:22
Don't write things like $\log_8 A - \log_8 B = \log_8\left(\frac{\log A}{\log B}\right)$; that's wrong. The identity should say $\log_8 A - \log_8 B = \log_8\left(\frac{A}{B}\right)$. – Michael Hardy Aug 20 '11 at 21:23
up vote 2 down vote accepted

Remember that $a \log x = \log x^a$.

Thus, we have $\log_8(5)-2\log_8(6) = \log_8(5)-\log_8(6^2) = \log_8(5)-\log_8(36)$

Now as you mentioned, since they have the same base you can apply the rule for differences between logarithms: $\log(x) -\log(y) = \log(\frac{x}{y})$

Thus we have $\log_8(5)-\log_8(36)= \log_8(\frac{5}{36})$.

Now to get back where you went wrong, you brought the 2 in front of $\log_8(6)$ inside by just multiplying, which is just wrong. The proper rule is: $a \log x = \log x^a$, which I mentioned in the beginning of the answer.

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Thank you for that very elaborative answer :) – John Aug 20 '11 at 21:09
You're welcome! – sxd Aug 20 '11 at 21:11

Remember that $2\;\log_8(6)=\log_8(6^2)$.

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so its 5/36 :D thnx m8 – John Aug 20 '11 at 21:07
Yes, $\log_8\left(\frac{5}{36}\right) = \log_8\left(\frac{5}{9}\right)-\frac{2}{3}$. – robjohn Aug 20 '11 at 21:11

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