If u and v are positive real numbers such that \frac{(u^{-2} v^3)^3}{u^k v^{-1}} = \frac{v… — ACT

If $u$ and $v$ are positive real numbers such that $\frac{(u^{-2} v^3)^3}{u^k v^{-1}} = \frac{v^{10}}{u^{10}}$ , what is the value of the exponent $k$ ?

Answer: 4

Answer

The value of the exponent is 4.

By applying the exponent rules systematically, the expression on the left simplifies to

u^{-6-k} v^{10}

, and the expression on the right is

u^{-10} v^{10}

. Equating the exponents of

u

gives

-6-k = -10

, which solves to

k = 4

Step-by-Step Solution

Apply the power of a product and power of a power properties to the numerator

(u^{-2} v^3)^3

u^{-6} v^9

According to the power of a product rule,

(xy)^a = x^a y^a

, and the power of a power rule,

(x^a)^b = x^{ab}

Use the quotient of powers property to simplify the left side of the equation.

u^{-6-k} v^{10}

The quotient of powers rule states that

\frac{x^a}{x^b} = x^{a-b}

, so the exponents of like bases are subtracted:

-6 - k

for

u

and

9 - (-1) = 10

for

v

Rewrite the right side of the equation,

\frac{v^{10}}{u^{10}}

, using a negative exponent.

u^{-10} v^{10}

Applying the negative exponent rule,

\frac{1}{x^a} = x^{-a}

Set the simplified expressions equal and solve for

k

k = 4

Since

u^{-6-k} v^{10} = u^{-10} v^{10}

, the exponents of the base

u

must be equal. Therefore,

-6 - k = -10

, which simplifies to

k = 4

Key Concept

Properties of Exponents in Algebraic Expressions

Estimated Time:1m 30s

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