Question

question 1 (1 point) solve $x = \log_{2}\left(\frac{1}{8}\
ight)$ by rewriting in exponential form. \bigcirc a) $\frac{1}{64}$ \bigcirc b) $-3$ \bigcirc c) $3$ \bigcirc d) $4$

Explanation:

Step1: Recall log to exponential rule

The logarithmic equation \( y = \log_b(a) \) can be rewritten in exponential form as \( b^y = a \). Here, \( b = 2 \), \( y = x \), and \( a=\frac{1}{8} \), so we get \( 2^x=\frac{1}{8} \).

Step2: Express \(\frac{1}{8}\) as a power of 2

We know that \( 8 = 2^3 \), so \( \frac{1}{8}=\frac{1}{2^3} \). Using the negative - exponent rule \( \frac{1}{a^n}=a^{-n} \), we can rewrite \( \frac{1}{2^3} \) as \( 2^{-3} \). So the equation \( 2^x=\frac{1}{8} \) becomes \( 2^x = 2^{-3} \).

Step3: Solve for x

If \( a^m=a^n \) (where \( a>0,a
eq1 \)), then \( m = n \). Since the base of both sides of the equation \( 2^x = 2^{-3} \) is 2 (and \( 2>0,2
eq1 \)), we can conclude that \( x=-3 \).

Answer:

b) -3