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

Question

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

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Allison 6 days 2021-09-10T10:23:13+00:00 2 Answers 0

Answers ( )

    0
    2021-09-10T10:24:45+00:00

    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

    0
    2021-09-10T10:25:12+00:00

    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

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