Question

rewrite each equation as requested.
(a) rewrite as a logarithmic equation.
$6^1 = 6$
$log_{\square}\square = \square$
(b) rewrite as an exponential equation.
$log_4 \frac{1}{64} = -3$
$\square^{\square} = \square$

Explanation:

Part (a)

Step1: Recall the exponential - logarithmic conversion formula

The general formula for converting an exponential equation \(a^{b}=c\) to a logarithmic equation is \(\log_{a}c = b\).

Step2: Apply the formula to \(6^{1}=6\)

Here, \(a = 6\), \(b = 1\), and \(c = 6\). Using the formula \(\log_{a}c=b\), we substitute the values: \(\log_{6}6 = 1\).

Part (b)

Step1: Recall the logarithmic - exponential conversion formula

The general formula for converting a logarithmic equation \(\log_{a}c=b\) to an exponential equation is \(a^{b}=c\).

Step2: Apply the formula to \(\log_{4}\frac{1}{64}=- 3\)

Here, \(a = 4\), \(b=-3\), and \(c=\frac{1}{64}\). Using the formula \(a^{b}=c\), we substitute the values: \(4^{-3}=\frac{1}{64}\).

Answer:

(a) \(\log_{\boldsymbol{6}}\boldsymbol{6}=\boldsymbol{1}\)
(b) \(\boldsymbol{4}^{\boldsymbol{-3}}=\boldsymbol{\frac{1}{64}}\)