Question

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

Answer 1

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:

$y = 0$

Answer 2

Answer:

Value of y=0

Step-by-step explanation:

We need to solve

$8^3*8^{-5}*8^y=1/8^2$

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

$8^3*8^{-5}*8^y=8^{-2}$

$8^y=\frac{8^{-2}}{8^3*8^{-5}}\\8^y=\frac{8^{-2}}{8^{3-5}}\\8^y=\frac{8^{-2}}{8^{-2}}\\8^y=1$

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