If x is a positive integer such that \frac{\sqrt{54 \cdot 4^{2x+1} + 9 \cdot 4^{2x}}}{\sqr… — GRE General Test

x

is a positive integer such that

\frac{\sqrt{54 \cdot 4^{2x+1} + 9 \cdot 4^{2x}}}{\sqrt{2^{x+3} + 2^x}} = 2560

what is the value of

x

Answer: 6

The value of

x

6

Factoring out common exponential terms inside both radicals yields

\sqrt{225 \cdot 2^{4x}} = 15 \cdot 2^{2x}

for the numerator and

\sqrt{9 \cdot 2^x} = 3 \cdot 2^{x/2}

for the denominator. Dividing these gives

5 \cdot 2^{3x/2}

. Equating this to

2560

results in

2^{3x/2} = 512 = 2^9

, which simplifies to

\frac{3x}{2} = 9

, or

x = 6

Simplify the numerator inside the radical expression.

\sqrt{54 \cdot 4^{2x+1} + 9 \cdot 4^{2x}} = 15 \cdot 2^{2x}

Rewrite

4^{2x+1}

4 \cdot 4^{2x}

. Then factor out

4^{2x}

54(4 \cdot 4^{2x}) + 9 \cdot 4^{2x} = (216 + 9)4^{2x} = 225 \cdot 4^{2x}

. Taking the square root gives

\sqrt{225} \cdot \sqrt{(2^2)^{2x}} = 15 \cdot 2^{2x}

Simplify the denominator inside the radical expression.

\sqrt{2^{x+3} + 2^x} = 3 \cdot 2^{x/2}

Rewrite

2^{x+3}

2^3 \cdot 2^x = 8 \cdot 2^x

. Factoring out

2^x

gives

(8 + 1)2^x = 9 \cdot 2^x

. Taking the square root gives

\sqrt{9} \cdot \sqrt{2^x} = 3 \cdot 2^{x/2}

Simplify the quotient of the two radical expressions.

\frac{15 \cdot 2^{2x}}{3 \cdot 2^{x/2}} = 5 \cdot 2^{3x/2}

Divide the constants

\frac{15}{3} = 5

and subtract exponents with the same base:

2x - \frac{x}{2} = \frac{3x}{2}

Equate to 2560 and solve for

x

x = 6

Divide both sides by 5:

2^{3x/2} = \frac{2560}{5} = 512

. Express 512 as a power of 2:

512 = 2^9

. Therefore,

\frac{3x}{2} = 9 \implies 3x = 18 \implies x = 6

Exponent and Radical Simplification using Base Prime Factorization

Estimated Time:2m 0s

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