Questions: Exponential and Logarithmic Functions Converting between logarithmic and exponential equat Rewrite each equation as requested. (a) Rewrite as an exponential equation. log2 32=5 (b) Rewrite as a logarithmic equation. 7^-2=1/49

Exponential and Logarithmic Functions
Converting between logarithmic and exponential equat

Rewrite each equation as requested.
(a) Rewrite as an exponential equation.
log2 32=5
(b) Rewrite as a logarithmic equation.
7^-2=1/49
Transcript text: Exponential and Logarithmic Functions Converting between logarithmic and exponential equat Rewrite each equation as requested. (a) Rewrite as an exponential equation. \[ \log _{2} 32=5 \] (b) Rewrite as a logarithmic equation. \[ 7^{-2}=\frac{1}{49} \]
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Solution

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Solution Steps

To solve the given problems, we need to convert between logarithmic and exponential forms.

(a) The logarithmic equation \(\log_{2} 32 = 5\) can be rewritten in exponential form. The general rule is that \(\log_{b} a = c\) is equivalent to \(b^c = a\).

(b) The exponential equation \(7^{-2} = \frac{1}{49}\) can be rewritten in logarithmic form. The general rule is that \(b^c = a\) is equivalent to \(\log_{b} a = c\).

Step 1: Convert Logarithmic to Exponential Form

Given the logarithmic equation: \[ \log_{2} 32 = 5 \] we can rewrite it in exponential form as: \[ 2^5 = 32 \] This confirms that the exponential form is correct.

Step 2: Convert Exponential to Logarithmic Form

Given the exponential equation: \[ 7^{-2} = \frac{1}{49} \] we can rewrite it in logarithmic form as: \[ \log_{7} \left(\frac{1}{49}\right) = -2 \] This shows that the logarithmic form is also correct.

Final Answer

The answers to the sub-questions are:

  • (a) The exponential form is \(2^5 = 32\).
  • (b) The logarithmic form is \(\log_{7} \left(\frac{1}{49}\right) = -2\).

Thus, the final answers are: \[ \boxed{2^5 = 32} \] \[ \boxed{\log_{7} \left(\frac{1}{49}\right) = -2} \]

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