What is y if 8^x = 2 and 3^(x+y) = 81?

You need to use the property of logarithms such that:
`log_a (b^c) = c*log_a b` , hence you should take logarithms to base 3 both sides
of `3^(x+y) = 81`  such that:

`log_3(3^(x+y)) = log_3 81
=gt (x+y)*log_3 3 = log_3 (3^4)`

Substituting 1 for `log_3
3`  yields:

`(x+y)*1 = 4*1 =gt x+y =
4`

Notice that you should find x using the exponential `8^x
= 2`  such that:

`8 = 2^3 =gt 8^x = (2^3)^x =gt 8^x =
2^(3x)`

Hence, you need to substitute `2^(3x)`  for `8^x `
in equation `8^x = 2`  such that:

`2^(3x) =` 
`2` 

Notice that bases of exponentials both sides are
alike, hence you should equate exponents such that:

`3x = 1
=gt x = 1/3`

You need to substitute `1/3`  for x in `x+y =
4`  such that:

`1/3 + y = 4 =gt y = 4 -
1/3`

Bringing the terms to a common denominator
yields:

`y = (12-1)/3 =gt y =
11/3`

Hence, evaluating y under
given conditions yields `y=11/3` .