If \frac{2^{x+3} \cdot 4^{2x-1}}{8^{x+1}} = 64, what is the value of (x + 2)^2? — State PSC Exam

If $\frac{2^{x+3} \cdot 4^{2x-1}}{8^{x+1}} = 64$ , what is the value of $(x + 2)^2$ ?

Cevap: 36

The value of

(x + 2)^2

is 36.

Converting all terms to base 2 simplifies the equation to

2^{2x-2} = 2^6

. Equating the powers gives

2x - 2 = 6

, so

x = 4

. Substituting

x = 4

into

(x + 2)^2

yields

(4 + 2)^2 = 36

Convert all exponential terms to base 2

Numerator term

4^{2x-1} = 2^{4x-2}

, denominator term

8^{x+1} = 2^{3x+3}

, and right-hand side

64 = 2^6

To combine powers using index laws, all expressions must share a common base.

Simplify the left-hand side expression using exponent laws

\frac{2^{x+3} \cdot 2^{4x-2}}{2^{3x+3}} = \frac{2^{5x+1}}{2^{3x+3}} = 2^{(5x+1)-(3x+3)} = 2^{2x-2}

Apply product law

a^m \cdot a^n = a^{m+n}

and quotient law

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

Solve for the variable x

2^{2x-2} = 2^6 \implies 2x - 2 = 6 \implies x = 4

When bases are equal, exponents must be equal.

Evaluate the requested target expression

(4 + 2)^2 = 6^2 = 36

Substitute the calculated value of

x = 4

into

(x + 2)^2

Laws of Indices and Exponential Equations

Tahmini Süre:1m 30s

Soru