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
(A)
(B)
(C)
(D) 2
1 Answer
C
Explanation:
From the formula for changing logarithm bases, you have
Dividing both sides by
Inverting both sides:
So, you have
The equation becomes
Using the property
The equation becomes
By the same logic as before, you have
Which, by befinition, means that