If x is a real number satisfying the equation \frac{\left(2^{3x+2} + 2^{3x}\right)^3}{\lef… — GRE General Test

x

is a real number satisfying the equation

\frac{\left(2^{3x+2} + 2^{3x}\right)^3}{\left(4^{2x+1} - 4^{2x}\right)^2} = \frac{2000}{9}

what is the value of

x

Answer: 4

Factoring out

2^{3x}

in the numerator gives

2^{3x}(2^2 + 1) = 5 \cdot 2^{3x}

. Cubing this yields

125 \cdot 2^{9x}

. In the denominator, rewriting

4^{2x}

2^{4x}

and factoring gives

2^{4x}(4 - 1) = 3 \cdot 2^{4x}

. Squaring this yields

9 \cdot 2^{8x}

. Taking the quotient gives

\frac{125}{9} \cdot 2^x

. Setting this equal to

\frac{2000}{9}

leads directly to

125 \cdot 2^x = 2000 \implies 2^x = 16 \implies x = 4

Factor out common exponential terms inside the parentheses

Numerator inside becomes

5 \cdot 2^{3x}

and denominator inside becomes

3 \cdot 2^{4x}

Factoring out

2^{3x}

from

2^{3x+2} + 2^{3x}

isolates the constant multiplier

(4+1)

, and expressing

4^{2x}

2^{4x}

allows base unification.

Raise the simplified terms to their respective outer powers

Numerator becomes

125 \cdot 2^{9x}

and denominator becomes

9 \cdot 2^{8x}

Using power rules

(a \cdot b)^n = a^n b^n

and

(a^m)^n = a^{m n}

Simplify the fraction by subtracting exponents of like bases

The left side simplifies to

\frac{125}{9} \cdot 2^x

\ me{2^{9x}}{2^{8x}} = 2^{9x-8x} = 2^x

using the quotient rule for exponents.

Solve the resulting single-variable exponential equation

2^x = 16

, which yields

x = 4

Multiplying both sides by

9

yields

125 \cdot 2^x = 2000

, so

2^x = 16 = 2^4

Factoring exponential expressions and applying power of a power and quotient rules

Question