(log₃13) (log₁₃x) (logₓy) = 2 Solve for y. ?

2 Answers
Mar 13, 2018

Since #log_3(13) = 1/(log_13(3))#
we have
#(log_3(13))(log_13(x))(log_x(y) ) = (log_13(x) /(log_13(3)))(log_x(y) )#
The quotient with a common base of 13 follows the change of base formula, so that
#log_13(x) /(log_13(3)) = log_3(x)#, and
the left hand side equals
#(log_3(x))(log_x(y))#

Since
#log_3(x) = 1/(log_x(3))#
the left side equals
#log_x(y)/log_x(3)#
which is a change of base for
#log_3(y)#

Now that we know that #log_3(y) = 2#, we convert to exponential form, so that
#y = 3^2 = 9#.

Mar 13, 2018

#y=9#

Explanation:

After using #log_a(b)*log(b)_c=log_a(c)# identity,

#log_3(13)*log_13(x)*log_x(y)=2#

#log_3(x)*log_x(y)=2#

#log_3(y)=2#

#y=3^2=9#