If x is a real number such that \frac{(3^x)^4 \cdot 9^{2-x}}{27^{x-1}} = 81, what is the v… — ACT

If $x$ is a real number such that $\frac{(3^x)^4 \cdot 9^{2-x}}{27^{x-1}} = 81$ , what is the value of $x$ ?

Answer: 3

The value of

x

is 3.

Rewriting the bases in terms of 3, the expression becomes

\frac{3^{4x} \cdot 3^{4-2x}}{3^{3x-3}} = 3^4

. Combining the numerator using the product rule gives

3^{2x+4}

in the numerator. Dividing by the denominator using the quotient rule gives

3^{(2x+4)-(3x-3)} = 3^{-x+7}

. Equating this to

3^4

gives

-x + 7 = 4

, which yields

x = 3

Express each base in terms of the common base 3

9 = 3^2

27 = 3^3

, and

81 = 3^4

Converting all terms to a common base allows the use of exponent rules to combine them.

Apply the power of a power rule

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

to rewrite each exponent

(3^x)^4 = 3^{4x}

(3^2)^{2-x} = 3^{4-2x}

, and

(3^3)^{x-1} = 3^{3x-3}

This simplifies individual terms by multiplying their exponents.

Apply the product rule of exponents

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

to combine the terms in the numerator

3^{4x} \cdot 3^{4-2x} = 3^{4x + 4 - 2x} = 3^{2x + 4}

Multiplying exponential terms with the same base is simplified by adding their exponents.

Apply the quotient rule of exponents

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

to simplify the fraction

3^{(2x+4) - (3x-3)} = 3^{-x+7}

Dividing exponential terms with the same base is simplified by subtracting the exponent in the denominator from the exponent in the numerator.

Equate the exponents of the simplified base 3 expression and base 3 representation of 81

-x + 7 = 4

, which solves to

x = 3

Since the bases are equal, the powers must be equal for the equation to hold true.

Properties of Exponents in Algebraic Expressions

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