For all real numbers x, what is the numerical value of the expression \frac{3^{x+2} - 3^x}… — GRE General Test

For all real numbers $x$ , what is the numerical value of the expression $\frac{3^{x+2} - 3^x}{3^{x-1} + 3^{x-2}}$ ?

Answer: 18

The numerical value of the expression is 18.

Factoring

3^x

from both terms in the numerator gives

3^x(3^2 - 1) = 8 \cdot 3^x

. Factoring

3^x

from both terms in the denominator gives

3^x(3^{-1} + 3^{-2}) = 3^x\left(\frac{1}{3} + \frac{1}{9}\right) = \frac{4}{9} \cdot 3^x

. Dividing the two expressions cancels

3^x

entirely, resulting in

\frac{8}{\frac{4}{9}} = 8 \times \frac{9}{4} = 18

Factor out

3^x

from the numerator.

3^{x+2} - 3^x = 3^x(3^2 - 1) = 3^x(9 - 1) = 8 \cdot 3^x

Applying the exponent rule

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

allows factoring out the common exponential factor

3^x

Factor out

3^x

from the denominator.

3^{x-1} + 3^{x-2} = 3^x(3^{-1} + 3^{-2}) = 3^x\left(\frac{1}{3} + \frac{1}{9}\right) = 3^x\left(\frac{4}{9}\right)

Applying negative exponent rules

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

allows evaluating the remaining numerical terms inside the parentheses.

Simplify the overall fraction by dividing the factored numerator by the factored denominator.

\frac{8 \cdot 3^x}{\frac{4}{9} \cdot 3^x} = \frac{8}{\frac{4}{9}} = 8 \times \frac{9}{4} = 18

The non-zero common term

3^x

cancels from both numerator and denominator, leaving a constant integer.

Factoring and simplifying exponential expressions with variable exponents.

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