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Evaluate the expression

$\log_8{8^{17}}$

I ended up getting $8^x = 8^{17}$.

I'm guessing I find x, but that's a huge number, and I feel like I'm doing this wrong.

user1176409
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Kyle
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    By inspection, what should $x$ be? – Amzoti Jan 21 '14 at 15:38
  • The log of a number means, what power do you have to raise the base by to get the thing that you are taking the logarithm of ... i.e. in your case, what power do you raise $8$ by to get $8^{17}$ ... – David Simmons Jan 21 '14 at 15:53

2 Answers2

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For any $0<a \neq 1$, $b>0$ and $r \in \mathbb{R}$, we had that:

$$ \textrm{log}_{a}b^{r} = r \cdot \textrm{log}_{a}b $$

In your example:

$$ \textrm{log}_{8}8^{17}=17 \cdot \textrm{log}_{8}8 $$

Como $\textrm{log}_{b}b=1 \; \Rightarrow \textrm{log}_{8}8=1$

And because of that we had:

$$ \textrm{log}_{8}8^{17}=17 \cdot \textrm{log}_{8}8=17 \cdot 1 = 17 $$

Décio Soares
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$x=\log_8{8^{17}}=\frac{\log{8^{17}}}{\log{8}}=17\frac{\log{8}}{\log{8}}=17$.

You use the property for changing the base of the logarithm at the second equality.

Kal S.
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