What is the value of 'a' for which (log_a 7)/(log_6 7) = log_pi 36 holds good, is ?

(A) 1/pi

(B) pi^2

(C) sqrtpi

(D) 2

1 Answer
Jun 13, 2018

C

Explanation:

From the formula for changing logarithm bases, you have

log_a(b) = log_c(b)/log_c(a)

Dividing both sides by log_c(b):

log_a(b)/log_c(b) = 1/log_c(a)

Inverting both sides:

log_c(b)/log_a(b) = log_c(a)

So, you have log_a(7)/log_6(7) = log_a(6)

The equation becomes

log_a(6) = log_pi(36)

Using the property log_a(b^c)=clog_a(b), you have

log_pi(36)=log_pi(6^2)=2log_pi(6)

The equation becomes

log_a(6) = 2log_pi(6)

therefore log_a(6)/log_pi(6) = 2

By the same logic as before, you have log_a(6)/log_pi(6) = log_a(pi). The equation becomes

log_a(pi) = 2

Which, by befinition, means that a^2=pi, thus a=sqrt(pi) (you can't use negative numbers as the base of a logarithm, so -sqrt(pi) is not an option).