Question

The value of $\frac{1}{{\log }_{3}12}+\frac{1}{{\log }_{4}12}$ is

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. $2$

Answer 1

The correct option is C

$1$

The explanation for the correct answer.

Find the value

Given: $\frac{1}{{\log }_{3}12}+\frac{1}{{\log }_{4}12}$

Formula ${\log }_{\mathrm{b}}\mathrm{a}=\frac{\mathrm{loga}}{\mathrm{logb}}$

$\begin{array}{rcl}\frac{1}{{\log }_{3}12}+\frac{1}{{\log }_{4}12}& =& \frac{1}{{\displaystyle \frac{\log 12}{\log 3}}}+\frac{1}{{\displaystyle \frac{\log 12}{\log 4}}}\\ & =& \frac{\log 3}{\log 12}+\frac{\log 4}{\log 12}\\ & =& \frac{\log 3+\log 4}{\log 12}\\ & =& \frac{\log \left(3\times 4\right)}{\log 12}\left[\mathrm{loga}+\mathrm{logb}=\log (\mathrm{a}\times \mathrm{b})\right]\\ & =& \frac{\log 12}{\log 12}\\ & =& 1\end{array}$

Hence $o\mathrm{ption}\left(\mathrm{C}\right)$ is the correct answer.