If k is a positive integer such that \sqrt{8^{10} + 2^{23} + 4^7} = 2^k + 128, what is the… — GMAT

If $k$ is a positive integer such that $\sqrt{8^{10} + 2^{23} + 4^7} = 2^k + 128$ , what is the value of $k$ ?

Answer: 15

The value of

k

is 15.

Converting all terms under the square root to base 2 produces

\sqrt{2^{30} + 2^{23} + 2^{14}}

. Recognizing that

(2^{15} + 2^7)^2 = (2^{15})^2 + 2(2^{15})(2^7) + (2^7)^2 = 2^{30} + 2^{23} + 2^{14}

, taking the square root yields

2^{15} + 2^7 = 2^{15} + 128

. Matching this with

2^k + 128

yields

k = 15

Convert terms under the square root to base 2.

The radical expression becomes

\sqrt{2^{30} + 2^{23} + 2^{14}}

Expressing terms with the same base allows exponent rules and algebraic identity recognition.

Identify the expression under the radical as a perfect square of the form

(a + b)^2

Setting

a = 2^{15}

and

b = 2^7

gives

2ab = 2 \cdot 2^{15} \cdot 2^7 = 2^{23}

, so

2^{30} + 2^{23} + 2^{14} = (2^{15} + 2^7)^2

The middle term

2^{23}

satisfies

2 \cdot \sqrt{2^{30}} \cdot \sqrt{2^{14}}

Evaluate the square root and solve for

k

\sqrt{(2^{15} + 2^7)^2} = 2^{15} + 128

. Setting

2^{15} + 128 = 2^k + 128

yields

k = 15

Comparing terms directly after evaluating

2^7 = 128

isolates

2^k = 2^{15}

Application of exponent rules combined with perfect square algebraic identities under radicals

Question