If x is a real number such that \sqrt[3]{\frac{9^{2x+1} \cdot 27^{x-1}}{3^{x+2}}} = 243, w… — GMAT

If $x$ is a real number such that $\sqrt[3]{\frac{9^{2x+1} \cdot 27^{x-1}}{3^{x+2}}} = 243$ , what is the value of $x$ ?

Answer: 3

The value of

x

is 3.

Converting all terms inside the radical to powers of 3 yields

\sqrt[3]{\frac{3^{4x+2} \cdot 3^{3x-3}}{3^{x+2}}} = \sqrt[3]{3^{6x-3}} = 3^{2x-1}

. Setting

3^{2x-1} = 243 = 3^5

gives

2x - 1 = 5

, which solves to

x = 3

Express all exponential terms inside the radical using a common base of 3.

9^{2x+1} = 3^{4x+2}

and

27^{x-1} = 3^{3x-3}

Unifying bases allows the application of exponent multiplication and division rules.

Simplify the algebraic expression inside the cube root.

\frac{3^{4x+2} \cdot 3^{3x-3}}{3^{x+2}} = 3^{(4x+2) + (3x-3) - (x+2)} = 3^{6x-3}

Adding exponents for multiplication of like bases and subtracting for division.

Simplify the radical expression using fractional exponent rules.

\sqrt[3]{3^{6x-3}} = (3^{6x-3})^{1/3} = 3^{2x-1}

Taking the

n

-th root of a power corresponds to dividing the exponent by

n

Solve for

x

by equating the exponent to the prime factorized form of 243.

3^{2x-1} = 3^5 \implies 2x - 1 = 5 \implies x = 3

Since the bases are identical and non-zero, the exponents must be equal.

Exponent and radical rules with common base conversion

Question