Question

1. for what real number value of x is the equation $64^{\frac{1}{3}} = 2^x$ true?

f. $\frac{1}{3}$
g. 2
h. 4
j. 6
k. 8

Explanation:

Step1: Express 64 as a power of 2

We know that \( 64 = 2^6 \), so we can rewrite the left - hand side of the equation \( 64^{\frac{1}{3}}=2^{x} \) as \( (2^6)^{\frac{1}{3}} \).

Step2: Use the power - of - a - power rule

The power - of - a - power rule states that \( (a^m)^n=a^{m\times n} \). For \( (2^6)^{\frac{1}{3}} \), we have \( m = 6 \) and \( n=\frac{1}{3} \). Then \( (2^6)^{\frac{1}{3}}=2^{6\times\frac{1}{3}} \).

Step3: Simplify the exponent

Calculate \( 6\times\frac{1}{3}=\frac{6}{3} = 2 \). So \( (2^6)^{\frac{1}{3}}=2^{2} \). Now our equation \( 64^{\frac{1}{3}}=2^{x} \) becomes \( 2^{2}=2^{x} \).

Step4: Solve for x

If two exponential functions with the same base (\( a^m=a^n \) where \( a>0,a
eq1 \)) are equal, then their exponents are equal. So since \( 2^{2}=2^{x} \), we have \( x = 2 \).

Answer:

G