Question

Sum of $21$ terms of series ${\log }_{9^{\frac{1}{2}}}\left(x\right)+{\log }_{9^{\frac{1}{3}}}\left(x\right)+{\log }_{9^{\frac{1}{4}}}\left(x\right).....$ is $252$, then the value of $x$ is:

• A. $7$
• B. $243$
• C. $9$
• D. $81$

Answer 1

The correct option is C

$9$

Explanation for the correct option:

Simplifying the series of $21$ terms:

The known logarithmic formula is ${\log }_b\left(x\right)=\frac{\log \left(a\right)}{\log \left(b\right)}$. Use this to simplify the series.

$$\begin{gathered} {\log }_{9^{\frac{1}{2}}}\left(x\right)+{\log }_{9^{\frac{1}{3}}}\left(x\right)+{\log }_{9^{\frac{1}{4}}}\left(x\right)+...+{\log }_{9^{\frac{1}{22}}}\left(x\right) \\ =\frac{\log \left(x\right)}{\log \left(9^{{\displaystyle \frac{1}{2}}}\right)}+\frac{\log \left(x\right)}{\log \left(9^{{\displaystyle \frac{1}{3}}}\right)}+\frac{\log \left(x\right)}{\log \left(9^{{\displaystyle \frac{1}{4}}}\right)}+...+\frac{\log \left(x\right)}{\log \left(9^{{\displaystyle \frac{1}{22}}}\right)} \\ =\frac{\log \left(x\right)}{{\displaystyle \frac{1}{2}\log \left(9\right)}}+\frac{\log \left(x\right)}{{\displaystyle \frac{1}{3}\log \left(9\right)}}+\frac{\log \left(x\right)}{{\displaystyle \frac{1}{4}\log \left(9\right)}}+...+\frac{\log \left(x\right)}{{\displaystyle \frac{1}{22}\log \left(9\right)}} \\ =\frac{2\log \left(x\right)}{\log \left(9\right)}+\frac{3\log \left(x\right)}{\log \left(9\right)}+\frac{4\log \left(x\right)}{\log \left(9\right)}+...+\frac{22\log \left(x\right)}{\log \left(9\right)} \\ =(2+3+4+...+22)\frac{\log \left(x\right)}{\log \left(9\right)} \\ =(2+3+4+...+22){\log }_9\left(x\right) \end{gathered}$$

The Sum of $n$ natural numbers is $\frac{n(n+1)}{2}$.

$$\begin{gathered} \Rightarrow (2+3+4+...+22){\log }_9\left(x\right)=(1+2+3+4+...+22-1){\log }_9\left(x\right) \\ =\left(\frac{22\times 23}{2}-1\right){\log }_9\left(x\right) \\ =252{\log }_9\left(x\right) \end{gathered}$$

The Sum of the series is given as $252$.

$$\begin{gathered} \Rightarrow 252{\log }_9\left(x\right)=252 \\ \Rightarrow {\log }_9\left(x\right)=1 \\ \Rightarrow \frac{\log \left(x\right)}{\log \left(9\right)}=1 \\ \Rightarrow \log \left(x\right)=\log \left(9\right) \\ \Rightarrow x=9 \end{gathered}$$

Therefore, option $\left(\mathrm{C}\right)$ is correct answer.