The problem asks to evaluate two expressions: a) $log_{\frac{1}{9}}27$ b) $5^{3log_5 20}$

1. Problem Description

The problem asks to evaluate two expressions:

a) $log_{\frac{1}{9}}27$

b) $5^{3log_5 20}$

2. Solution Steps

a) We want to find $x$ such that $(\frac{1}{9})^x = 27$.

We can rewrite $\frac{1}{9}$ as $9^{-1} = (3^2)^{-1} = 3^{-2}$ and $27$ as $3^3$.

Thus, we have $(3^{-2})^x = 3^3$, which means $3^{-2x} = 3^3$.

Therefore, $-2x = 3$, so $x = -\frac{3}{2}$.

b) Using the property $a^{log_a x} = x$, we want to rewrite the expression $5^{3log_5 20}$.

We can use the logarithm power rule $n log_a x = log_a (x^n)$ to rewrite the exponent.

$3log_5 20 = log_5 (20^3) = log_5 8000$.

Then $5^{3log_5 20} = 5^{log_5 8000} = 8000$.

3. Final Answer

a) $-\frac{3}{2}$

b) $8000$