If x is a positive integer such that \sqrt{\frac{8^{10} + 4^{10}}{8^4 + 4^{11}}} = 2^x, wh… — GMAT

If $x$ is a positive integer such that $\sqrt{\frac{8^{10} + 4^{10}}{8^4 + 4^{11}}} = 2^x$ , what is the value of $x$ ?

Answer: 4

Converting all terms to base 2 yields

\sqrt{\frac{2^{30}+2^{20}}{2^{12}+2^{22}}}

. Factoring out

2^{20}

in the numerator and

2^{12}

in the denominator leaves

\sqrt{\frac{2^{20}(2^{10}+1)}{2^{12}(2^{10}+1)}} = \sqrt{2^8} = 2^4

. Thus

2^x = 2^4

, so

x = 4

Express all terms with prime base 2 using exponent rules.

8^{10} = 2^{30}

4^{10} = 2^{20}

8^4 = 2^{12}

, and

4^{11} = 2^{22}

Converting non-prime bases to a common base enables exponent simplification and factoring.

Factor the numerator and denominator by pulling out the lowest power of 2 in each.

Numerator:

2^{20}(2^{10} + 1)

; Denominator:

2^{12}(1 + 2^{10})

Factoring isolates common terms in sums of powers so they can be canceled.

Cancel the identical factor

(2^{10} + 1)

and compute the radical expression.

\sqrt{\frac{2^{20}}{2^{12}}} = \sqrt{2^8} = 2^4 = 16.

Applying quotient rule for exponents

\frac{2^{20}}{2^{12}} = 2^{8}

and radical rule

\sqrt{2^8} = (2^8)^{1/2} = 2^4

Set

2^4

equal to

2^x

to solve for

x

x = 4.

Since the bases are identical and positive, the exponents must be equal.

Exponents, Roots, and Powers of Integers

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