Question

If $a^2+4b^2=12ab,$ then $\log (a+2b)$ is

• A. $\frac{1}{2}(4\log 2+\log a+\log b)$
• B. $0$
• C. $(\log 2+\log a+\log b)$
• D. $(4\log 2+\log a–\log b)$

Answer 1

The correct option is A

$\frac{1}{2}(4\log 2+\log a+\log b)$

Explanation for the correct option:

Find the value of $\mathit{l}\mathit{o}\mathit{g}\mathbf{(}\mathit{a}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathit{b}\mathbf{)}$:

Given $a^2+4b^2=12ab,$

Adding $4ab$on both sides, we get

$$\begin{gathered} a^2+4b^2+4ab=16ab \\ \Rightarrow {(a+2b)}^2=16ab \end{gathered}$$

Taking log on both sides

$$\begin{gathered} \log {(a+2b)}^2=\log \left(16ab\right) \\ \Rightarrow 2\log (a+2b)=\log 16+\log a+\log b \\ \Rightarrow 2\log (a+2b)=\log {\left(2\right)}^4+\log a+\log b \\ \Rightarrow 2\log (a+2b)=4\log 2+\log a+\log b \\ \Rightarrow \log (a+2b)=\frac{1}{2}(4\log 2+\log a+\log b) \end{gathered}$$

Hence, the correct option is (A).