# How do you evaluate ln(ln e^(e^100))?

Mar 31, 2016

100

#### Explanation:

$\ln \left({e}^{a}\right) = a$
The bracketed logarithm $\ln \left({\left(e\right)}^{{e}^{100}}\right) = {e}^{100}$
The given expression = $\ln \left({e}^{100}\right)$ = 100.

Apr 5, 2016

$100$

#### Explanation:

We have:

$\ln \left(\ln \left({e}^{{e}^{100}}\right)\right)$

Within the innermost logarithm, we can use the following rule:

$\ln \left({\textcolor{b l u e}{a}}^{\textcolor{red}{b}}\right) = \textcolor{red}{b} \cdot \ln \left(\textcolor{b l u e}{a}\right)$

This gives us:

$\ln \left(\ln \left({\textcolor{b l u e}{e}}^{\textcolor{red}{{e}^{100}}}\right)\right) = \ln \left(\textcolor{red}{{e}^{100}} \cdot \ln \left(\textcolor{b l u e}{e}\right)\right)$

Since $\ln \left(e\right) = 1$, this equals

$\ln \left({e}^{100} \cdot \ln \left(e\right)\right) = \ln \left({e}^{100}\right)$

Using the previously defined exponent rule, we can rewrite this as follows:

$\ln \left({\textcolor{b l u e}{e}}^{\textcolor{red}{100}}\right) = \textcolor{red}{100} \cdot \ln \left(\textcolor{b l u e}{e}\right) = \overline{\underline{|}} \textcolor{w h i t e}{\frac{a}{a}} 100 \textcolor{w h i t e}{\frac{a}{a}} |$