Discussion

Explanation:

Let log₅ 2 be rational.

Then log₅ 2 = $\frac{m}{n}$

∴ $5^{\frac{m}{n}}$ = 2

∴ ${\left(5^{m/n}\right)}^n$ = $2^n$

∴ 5^m = 2ⁿ

However, number 2 raised to any positive integer power must be even, but 5 raised to any positive integer power must be odd.

Hence, we have a contradiction.

∴ log₅ 2 is irrational.

Hence, option (d).

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