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
KillerBunny
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).

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