Question

which is equivalent to $3\log_{2}8 + 4\log_{2}\frac{1}{2} - \log_{3}2$?

$\bigcirc$ $5 - \log_{3}2$

$\bigcirc$ $16 - \log_{3}2$

$\bigcirc$ $\log_{2}48 - \log_{3}2$

$\bigcirc$ $\log_{3}32$

Explanation:

Step1: Simplify \( 3\log_{2}8 \)

We know that \( 8 = 2^3 \), so using the logarithm power rule \( \log_a a^b = b \), we have \( \log_{2}8=\log_{2}2^3 = 3 \). Then \( 3\log_{2}8 = 3\times3=9 \).

Step2: Simplify \( 4\log_{2}\frac{1}{2} \)

We know that \( \frac{1}{2}=2^{-1} \), so \( \log_{2}\frac{1}{2}=\log_{2}2^{-1}=-1 \). Then \( 4\log_{2}\frac{1}{2}=4\times(-1)= - 4 \).

Step3: Combine the first two terms

Now, add the results of the first two steps: \( 3\log_{2}8 + 4\log_{2}\frac{1}{2}=9+( - 4)=5 \).

Step4: Include the last term

The original expression is \( 3\log_{2}8 + 4\log_{2}\frac{1}{2}-\log_{3}2 \), so substituting the result from step 3, we get \( 5-\log_{3}2 \).

Answer:

\( 5 - \log_{3}2 \) (corresponding to the first option: \( 5-\log_{3}2 \))