TMUA practice: Exponentials Logs, problem 2

Place each number relative to $1$.

• $\;\log_{3} 2\;$: the base is $3$ and the argument is $2$. Since $\log_{3} 3 = 1$ and the logarithm to base $3$ is strictly increasing, $2 < 3$ gives $\log_{3} 2 < \log_{3} 3 = 1$. Also $1 < 2$ gives $\log_{3} 2 > \log_{3} 1 = 0$. So $\;0 < \log_{3} 2 < 1$.
• $\;\log_{2} 3\;$: the base is $2$ and the argument is $3$. Since $\log_{2} 2 = 1$ and $3 > 2$, the increasing function $\log_{2}$ gives $\log_{2} 3 > 1$.

Combining: $\;\log_{3} 2 < 1 < \log_{2} 3$, so $\log_{3} 2 < \log_{2} 3$. This is option A.

Alternative route. Use the change-of-base identity $\;\log_{2} 3 = 1 / \log_{3} 2$. Set $t = \log_{3} 2$; then $0 < t < 1$ from the first argument, and $\log_{2} 3 = 1/t$. Since $0 < t < 1$ implies $1/t > 1 > t$, we have $\log_{2} 3 > \log_{3} 2$. (As a bonus, this gives the well-known identity $\;\log_{3} 2 \cdot \log_{2} 3 = 1$.)

Why the other options fail.

• B is wrong by the strict bounds just established: one is below $1$, the other above.
• C reverses the correct inequality.
• D would require $\log_{2} 3 = -\log_{3} 2$, but both numbers are positive, so their sum cannot be $0$.
• E would require one factor to be negative. Both logarithms here are positive (their arguments are greater than $1$ and so are their bases), so their product is positive, not negative.

The lesson: to compare two logarithms, bracket each between known integer values (often $0$ and $1$, or $1$ and $2$) by comparing the argument to the base.