Respuesta :

carlosego

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:

[tex]a ^ n * a ^ m = a ^ {n + m}[/tex]

So, we can rewrite the given expression as:

[tex]8 ^ {3-5 + y} = \frac {1} {8 ^ 2}\\8 ^ {- 2 + y} = \frac {1} {8 ^ 2}[/tex]

So, if [tex]y = 0[/tex]:

[tex]8 ^ {- 2} = \frac {1} {8 ^ 2}\\\frac {1} {8 ^ 2} = \frac {1} {8 ^ 2}[/tex]

Equality is met!

Answer:

[tex]y = 0[/tex]

absor201

Answer:

Value of y=0

Step-by-step explanation:

We need to solve

[tex]8^3*8^{-5}*8^y=1/8^2[/tex]

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

[tex]8^3*8^{-5}*8^y=8^{-2}[/tex]

[tex]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[/tex]

Taking ln on both sides

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

So, value of y=0

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