Question

The value of $x$ which satisfies ${\log }_{3}x+{\log }_{9}x+{\log }_{27}x=\frac{11}{2}$ is equal to

• A. $27$
• B. $9$
• C. $3$
• D. $1$

Answer 1

The correct option is A $27$

${\log }_{3}x+{\log }_{9}x+{\log }_{27}x=\frac{11}{2}$

By using change of base rule, we get

$\Rightarrow \frac{\log x}{\log 3}+\frac{\log x}{\log 9}+\frac{\log x}{\log 27}=\frac{11}{2}$

$\Rightarrow \frac{\log x}{\log 3}+\frac{\log x}{\log 3^2}+\frac{\log x}{\log 3^3}=\frac{11}{2}$

$\Rightarrow \frac{\log x}{\log 3}+\frac{\log x}{2\log 3}+\frac{\log x}{3\log 3}=\frac{11}{2}$

$\Rightarrow \frac{\log x}{\log 3}(1+\frac{1}{2}+\frac{1}{3})=\frac{11}{2}$

$\Rightarrow \log x=\log 27$

Removing $\log$, we get

$\therefore x=27$

Hence, option A.