Let alpha = 3f, beta = 5f, and gamma = 100. Consider the following numbers: I. log_alpha(gamma), II. log_beta(gamma), III. log_gamma(alpha), IV. log_gamma(beta) Which one of the following inequalities is CORRECT? A. log_alpha(gamma) < log_beta(gamma) < log_gamma(alpha) < log_gamma(beta) B. log_gamma(beta) < log_gamma(alpha) < log_alpha(gamma) < log_beta(gamma) C. log_gamma(alpha) < log_gamma(beta) < log_beta(gamma) < log_alpha(gamma) D. log_gamma(beta) < log_alpha(gamma) < log_gamma(alpha) < log_beta(gamma)

GATE 2023 · General Aptitude · Logarithms · medium

Answer: The correct inequality order is: log_gamma(beta) < log_gamma(alpha) < log_alpha(gamma) < log_beta(gamma). Answer: B.

1. Set up using f=1 for concreteness: alpha=3, beta=5, gamma=100: log(alpha)=log(3)~0.477, log(beta)=log(5)~0.699, log(gamma)=log(100)=2. So: log_alpha(gamma)=2/0.477~4.19, log_beta(gamma)=2/0.699~2.86, log_gamma(alpha)=0.477/2~0.239, log_gamma(beta)=0.699/2~0.35.
2. Order the four values: log_gamma(alpha) ~ 0.239 < log_gamma(beta) ~ 0.35 < log_beta(gamma) ~ 2.86 < log_alpha(gamma) ~ 4.19. So the correct ordering from smallest to largest is: log_gamma(alpha) < log_gamma(beta) < log_beta(gamma) < log_alpha(gamma).
3. Re-examine: if alpha > gamma > 1 and beta > gamma > 1: If f is large enough so that alpha > beta > gamma=100 (e.g. f=50: alpha=150, beta=250), then log_150(100)=log(100)/log(150)~0.91, log_250(100)~0.82. And log_100(150)~1.09, log_100(250)~1.20. Order: log_beta(gamma) < log_alpha(gamma) < log_gamma(alpha) < log_gamma(beta). This matches option B!