Question

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

Answer 1

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`.

Answer 2

`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`