Question

8^3*8^-5*8^y=1/8^2, what is the value of y in the product of powers below?

Answer 1

Value of y=0

Explanation:

We need to solve

`8^3*8^(-5)*8^y=1/8^2`

We know that 1/a^2 = a^-2

`8^3*8^(-5)*8^y=8^(-2)`

`8^y=(8^(-2))/(8^3*8^(-5))\\8^y=(8^(-2))/(8^(3-5))\\8^y=(8^(-2))/(8^(-2))\\8^y=1`

Taking ln on both sides

`ln(8^y)=ln(1)\\yln(8)=ln(1)\\y= ln(1)/ln(8)\\We\,\,know\,that\,\,ln(1) =0\\y=0`

So, value of y=0

Answer 2

For this case we have that by definition of multiplication of powers of the same base, the same base is placed and the exponents are added:

`a ^ n * a ^ m = a ^ {n + m}`

So, we can rewrite the given expression as:

`8 ^ {3-5 + y} = \frac {1} {8 ^ 2}\\8 ^ {- 2 + y} = \frac {1} {8 ^ 2}`

So, if
`y = 0`:

`8 ^ {- 2} = \frac {1} {8 ^ 2}\\\frac {1} {8 ^ 2} = \frac {1} {8 ^ 2}`

Equality is met!

Answer:

`y = 0`