(a) e ^ -2ln5

(b)ln(ln e^e^10)

The second one is e to the e to the tenth power.

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- Sep 9th 2008, 04:29 PM #1

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- Sep 9th 2008, 04:37 PM #2
For a) Do you remember the rule $\displaystyle \log_a(m^p)=p\log_a(m)$?

Using this rule, we get

$\displaystyle e^{-2\ln(5)}=e^{\ln(5^{-2})}=5^{-2}=\frac{1}{5^2}=\frac{1}{25}$

The second actually just reduces to 10, because $\displaystyle \ln (e^{e^{10}})=e^{10}$ and so

$\displaystyle \ln[\ln(e^{e^{10}})]=\ln[e^{10}]=10$.

- Sep 9th 2008, 04:40 PM #3

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- Sep 9th 2008, 04:42 PM #4

- Sep 9th 2008, 04:56 PM #5

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- Sep 9th 2008, 04:58 PM #6

- Sep 9th 2008, 05:00 PM #7

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- Sep 9th 2008, 05:04 PM #8