If x is a real number such that \sqrt{\frac{4^{x+2} + 4^{x+2} + 4^{x+2} + 4^{x+2}}{8^{x-3}… — ACT

If $x$ is a real number such that $\sqrt{\frac{4^{x+2} + 4^{x+2} + 4^{x+2} + 4^{x+2}}{8^{x-3}}} = 64$ , what is the value of $x$ ?

Answer: 3

The value of

x

3

By rewriting the repeated addition in the numerator as

4 \times 4^{x+2} = 4^{x+3}

and converting all bases to 2, the expression inside the square root simplifies to

2^{15-x}

. Taking the square root gives

2^{\frac{15-x}{2}}

. Setting this equal to

64

(which is

2^6

) and equating the exponents yields

x = 3

Rewrite the sum in the numerator

4^{x+2} + 4^{x+2} + 4^{x+2} + 4^{x+2}

as a product.

4 \times 4^{x+2} = 4^{x+3}

Adding a term to itself four times is equivalent to multiplying that term by 4.

Convert the base 4 numerator and base 8 denominator to base 2.

Numerator:

(2^2)^{x+3} = 2^{2x+6}

; Denominator:

(2^3)^{x-3} = 2^{3x-9}

Expressing terms with the same base allows the use of exponent laws to simplify the fraction.

Simplify the fraction by subtracting the denominator's exponent from the numerator's exponent.

\frac{2^{2x+6}}{2^{3x-9}} = 2^{(2x+6)-(3x-9)} = 2^{15-x}

The quotient rule for exponents states that

\frac{a^m}{a^n} = a^{m-n}

Apply the square root to the simplified fraction, set it equal to

64

, and express both sides as powers of 2.

\sqrt{2^{15-x}} = 2^{\frac{15-x}{2}} = 64 = 2^6

The square root of a term is equivalent to raising that term to the power of

\frac{1}{2}

Equate the exponents and solve for

x

\frac{15-x}{2} = 6 \implies 15-x = 12 \implies x = 3

If two exponential expressions with the same positive base are equal, their exponents must be equal.

Simplifying expressions using the laws of exponents and properties of roots

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