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

2 Answers
Mar 13, 2018

Since log_3(13) = 1/(log_13(3))log3(13)=1log13(3)
we have
(log_3(13))(log_13(x))(log_x(y) ) = (log_13(x) /(log_13(3)))(log_x(y) )(log3(13))(log13(x))(logx(y))=(log13(x)log13(3))(logx(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)log13(x)log13(3)=log3(x), and
the left hand side equals
(log_3(x))(log_x(y))(log3(x))(logx(y))

Since
log_3(x) = 1/(log_x(3))log3(x)=1logx(3)
the left side equals
log_x(y)/log_x(3)logx(y)logx(3)
which is a change of base for
log_3(y)log3(y)

Now that we know that log_3(y) = 2log3(y)=2, we convert to exponential form, so that
y = 3^2 = 9y=32=9.

Mar 13, 2018

y=9y=9

Explanation:

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

log_3(13)*log_13(x)*log_x(y)=2log3(13)log13(x)logx(y)=2

log_3(x)*log_x(y)=2log3(x)logx(y)=2

log_3(y)=2log3(y)=2

y=3^2=9y=32=9