Answer
See explanation
Work Step by Step
Below is a standard proof of the logarithm power rule:
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## Theorem
For any real numbers \(a, b, x\) with \(b > 0\), \(b \neq 1\), and \(x > 0\), the following holds:
\[
\log_{b}(x^{a}) \;=\; a \,\log_{b}(x).
\]
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## Proof
1. **Recall the definition of \(\log_{b}(x)\):**
By definition, \(\log_{b}(x)\) is the exponent to which we raise \(b\) to get \(x\). Symbolically,
\[
b^{\log_{b}(x)} \;=\; x.
\]
2. **Express \(x^a\) using this definition:**
Since \(x = b^{\log_{b}(x)}\), it follows that
\[
x^{a}
\;=\;
\bigl(b^{\log_{b}(x)}\bigr)^{a}
\;=\;
b^{\,a \,\log_{b}(x)}.
\]
3. **Relate to \(\log_{b}(x^a)\):**
By definition of logarithm again,
\[
\log_{b}\bigl(x^a\bigr)
\quad\text{is the unique real number }y\text{ such that }b^{y} = x^a.
\]
But from Step 2, we already have \(x^a = b^{\,a \,\log_{b}(x)}\). So
\[
b^{y} \;=\; b^{\,a \,\log_{b}(x)}.
\]
4. **Use injectivity of the exponential function:**
Because \(b^u = b^v\) implies \(u = v\) for \(b>0\) and \(b\neq1\), it follows that
\[
y \;=\; a \,\log_{b}(x).
\]
But \(y\) was defined to be \(\log_{b}(x^a)\). Therefore,
\[
\log_{b}\bigl(x^a\bigr) \;=\; a \,\log_{b}(x).
\]
This completes the proof.
