Answer
See explanation
Work Step by Step
Below is a concise proof that \(\log_{3}(7)\) is irrational, using the same contradiction and unique factorization argument:
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### Proof that \(\log_{3}(7)\) is irrational
1. **Assume \(\log_{3}(7)\) is rational.**
Suppose, for contradiction, that
\[
\log_{3}(7) \;=\; \frac{p}{q},
\]
where \(p\) and \(q\) are integers with \(\gcd(p,q) = 1\) and \(q \neq 0\).
2. **Convert the logarithmic equation into an exponential equation.**
By the definition of logarithm,
\[
3^{p/q} = 7.
\]
Raising both sides to the power \(q\) gives
\[
3^{p} = 7^{q}.
\]
3. **Compare prime factorizations.**
- The left-hand side \(3^p\) has the prime factorization consisting solely of the prime \(3\) (repeated \(p\) times).
- The right-hand side \(7^q\) has the prime factorization consisting solely of the prime \(7\) (repeated \(q\) times).
4. **Derive a contradiction via the Fundamental Theorem of Arithmetic.**
If \(3^p = 7^q\), then the prime factorizations on both sides must match. This would require the prime \(3\) to be the same as the prime \(7\), which is impossible.
5. **Conclusion.**
The contradiction shows our assumption (that \(\log_{3}(7)\) is rational) is false. Therefore,
\[
\log_{3}(7) \text{ is irrational.}
\]
