If a is a real number greater than 1 such that a^x - a^{-x} = 2, what is the value of the … — GRE General Test

If $a$ is a real number greater than $1$ such that $a^x - a^{-x} = 2$ , what is the value of the expression $\frac{a^{3x} + a^{-3x}}{a^{2x} + a^{-2x}}$ ?

$\frac{5\sqrt{2}}{3}$ Answer
B
$\frac{4}{3}$
C
$2$
D
$\frac{5\sqrt{6}}{6}$
E
$\frac{5}{3}$

Answer

\frac{5\sqrt{2}}{3}

Squaring the given relationship

a^x - a^{-x} = 2

gives

a^{2x} - 2 + a^{-2x} = 4

, which simplifies to

a^{2x} + a^{-2x} = 6

. Adding 4 to both sides gives

(a^x + a^{-x})^2 = 8

, so

a^x + a^{-x} = 2\sqrt{2}

. By the sum of cubes identity,

a^{3x} + a^{-3x} = (a^x + a^{-x})(a^{2x} + a^{-2x} - 1) = 2\sqrt{2}(6 - 1) = 10\sqrt{2}

. Dividing

10\sqrt{2}

6

gives the simplified result

\frac{5\sqrt{2}}{3}

Step-by-Step Solution

Square both sides of the given equation

a^x - a^{-x} = 2

(a^x - a^{-x})^2 = a^{2x} - 2(a^x)(a^{-x}) + a^{-2x} = 4 \implies a^{2x} + a^{-2x} = 6

Expanding the binomial square allows us to find the denominator

a^{2x} + a^{-2x}

directly.

Determine the value of

a^x + a^{-x}

(a^x + a^{-x})^2 = a^{2x} + 2 + a^{-2x} = 6 + 2 = 8 \implies a^x + a^{-x} = \sqrt{8} = 2\sqrt{2}

Since

a > 1

a^x > 0

and

a^{-x} > 0

, their sum must be positive.

Use the sum of cubes identity

u^3 + v^3 = (u + v)(u^2 - uv + v^2)

to evaluate

a^{3x} + a^{-3x}

a^{3x} + a^{-3x} = (a^x + a^{-x})(a^{2x} - 1 + a^{-2x}) = (2\sqrt{2})(6 - 1) = 10\sqrt{2}

Factoring the numerator breaks it into terms whose numerical values are known.

Compute the ratio of numerator to denominator.

\frac{a^{3x} + a^{-3x}}{a^{2x} + a^{-2x}} = \frac{10\sqrt{2}}{6} = \frac{5\sqrt{2}}{3}.

Simplify the fraction by dividing numerator and denominator by 2.

Key Concept

Algebraic transformations of exponential expressions using polynomial identities.

Estimated Time:2m 30s

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