which is equivalent to $3\\log_28 + 4\\log_2\\frac{1}{2}-\\log_32$?\n$5 - \\log_32$\n$16…
which is equivalent to $3\\log_28 + 4\\log_2\\frac{1}{2}-\\log_32$?\n$5 - \\log_32$\n$16 - \\log_32$\n$\\log_248 - \\log_32$\n$\\log_332$\ndone
Answer
Explanation:
Step1: Simplify $3\log_28$
Use the property $n\log_aM=\log_aM^n$ and $\log_28 = \log_22^3=3$. So $3\log_28=3\times3 = 9$.
Step2: Simplify $4\log_2\frac{1}{2}$
Since $4\log_2\frac{1}{2}=4\log_22^{- 1}$, and using $n\log_aM=\log_aM^n$, we get $4\log_22^{-1}=4\times(-1)= - 4$.
Step3: Combine the first - two terms
$3\log_28 + 4\log_2\frac{1}{2}=9+( - 4)=5$.
Step4: Analyze the whole expression
The original expression $3\log_28 + 4\log_2\frac{1}{2}-\log_32$ simplifies to $5-\log_32$.
Answer:
$5-\log_32$