Question

find p such that: {(3)^-2}^p=81

Answer 1

Exponential and Logarithmic Equations

Answer:

$p = -2$

Step-by-step explanation:

Given:

$(3^{-2})^p = 81$

Recall:

$a^b = c \Leftrightarrow \log_a (c) = b$

$\log_a (b^c) = d \Leftrightarrow c \cdot \log_a (b) = d$

Solving for $p$:

$(3^{-2})^p = 81 \\ (\frac{1}{3^2})^p = 81 \\ (\frac{1}{9})^p = 81 \\ \log_\frac{1}{9} (81) = p \\ p = \log_\frac{1}{9} ((\frac{1}{81})^{-1}) \\ p = -\log_\frac{1}{9} (\frac{1}{81}) \\ p = -(2) \\ p = -2$