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Can you guys show me how you would solve this:

logn n^3 =

log2 4 =

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Two basic facts about logarithms are: $\log_n(n) = 1$ and $\log(a^b) = b\log(a)$. So $\log_n(n^a) = a$. –  Eelvex Mar 31 '11 at 3:51
Why would you call this an algorithm? –  picakhu Mar 31 '11 at 13:33
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Well, what does logarithm mean? $\log_n a$ means the number $x$ such that $n^x = a$.

In your first case $\log_n n^3$ is the number $x$ such that $n^x = n^3$. What is $x$? In your second case, $\log_2 4$ is the number $y$ such that $2^y = 4$, which I think you can find quickly.

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