For all non-zero real numbers y, the expression \frac{(y^3)^2 \cdot (y^{-4})^a}{y^5} is eq… — ACT

For all non-zero real numbers

y

, the expression

\frac{(y^3)^2 \cdot (y^{-4})^a}{y^5}

is equivalent to

y^{-7}

. What is the value of

a

The correct value of

a

is 2.

Applying the exponent rules systematically allows us to simplify the expression. First,

(y^3)^2

becomes

y^6

and

(y^{-4})^a

becomes

y^{-4a}

using the power of a power rule. Second, we combine the terms in the numerator using the product rule to get

y^{6-4a}

. Third, we divide by

y^5

using the quotient rule to obtain

y^{6-4a-5} = y^{1-4a}

. Setting this equal to the target expression

y^{-7}

gives the equation

1-4a = -7

. Solving for

a

gives

-4a = -8

, which simplifies to

2

Apply the power of a power rule,

(x^m)^n = x^{mn}

, to the exponential terms in the numerator.

(y^3)^2 = y^6

and

(y^{-4})^a = y^{-4a}

This simplifies nested exponent terms into single base terms.

Apply the product rule of exponents,

x^m \cdot x^n = x^{m+n}

, to combine the numerator terms.

y^6 \cdot y^{-4a} = y^{6-4a}

This simplifies the numerator to a single power of

y

Apply the quotient rule of exponents,

\frac{x^m}{x^n} = x^{m-n}

, to divide by the denominator.

\frac{y^{6-4a}}{y^5} = y^{(6-4a) - 5} = y^{1-4a}

This simplifies the entire rational expression into a single exponential expression.

Equate the simplified exponent to the exponent of the equivalent expression and solve the linear equation for

a

1-4a = -7 \implies -4a = -8 \implies a = 2

Since the bases are equal and non-zero, their exponents must be equal for the expressions to be equivalent.

Properties of Exponents in Algebraic Expressions

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