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` .**

You need to use property of logarithms such that: `log_a (b^c) = c*log_a b` , hence you should take logarithm 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) = 4 `

Notice that you may find x from the equation such that:

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

Substituting `2^(3x)` for 8^x in `8^x = 2 ` yields:

`2^(3x) = 2`

Notice that bases of exponentials are alike, hence the exponents are equal `3x = 1=gt x = 1/3` .

Substituting `1/3` for x in `(x+y) = 4` such that:

`1/3 + y = 4 `

`y = 4 - 1/3`

You need to bring the terms to a common denominator such that:

`y = (4*3 - 1)/3 `

`y = 11/3`

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

It is given that 8^x = 2 and 3^(x+y) = 81.

8^x = 2

=> (2^3)^x = 2^1

=> 2^(3*x) = 2^1

as the base is the same, we can equate the exponents

=> 3*x = 1

=> x = 1/3

3^(x + y) = 81

=> 3^(x + y) = 3^4

Again equate the exponents

=> x + y = 4

substitute x = 1/3

=> 1/3 + y = 4

=> y = 4 - 1/3

=> y = 11/3

**The required value of y = 11/3**

2^3x = 2

Since the bases are matching now, we'll apply one to one rule:3x=1 => x = 1/3

We'll consider the second exponential equation and we'll put 81 = 3^4

3^(x+y)= 3^4

Since the bases are matching now, we'll apply one to one rule:

x+y=4But x = 1/3 => 1/3 + y = 4 => y = 4 - 1/3 => y= 11/3

**The requested value of y is: y = 11/3.**