TAOCP 1.2.2 Exercise 18

Section 1.2.2: Numbers, Powers, and Logarithms

Exercise 18. [10] Prove or disprove: $\log_8 x = \frac{1}{2} \lg x$.

Verified: yes
Solve time: 26s

The statement is true. By the change-of-base formula (14),

$\log_8 x = \frac{\log_2 x}{\log_2 8}.$

Since $8 = 2^3$, Eq. (12) gives

$\log_2 8 = \log_2 (2^3) = 3 \log_2 2 = 3.$

Therefore

$\log_8 x = \frac{\log_2 x}{3} = \frac{1}{3} \lg x.$

Hence $\log_8 x \ne \frac{1}{2} \lg x$ in general. The proposition is disproved. ∎