Exponential and Logarithmic Equations
Answer:
[tex]p = -2[/tex]
Step-by-step explanation:
Given:
[tex](3^{-2})^p = 81[/tex]
Recall:
[tex]a^b = c \Leftrightarrow \log_a (c) = b[/tex]
[tex]\log_a (b^c) = d \Leftrightarrow c \cdot \log_a (b) = d[/tex]
Solving for [tex]p[/tex]:
[tex](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[/tex]
