Question

Solve for x
$log_{3}x + log_{9}x + log_{81}x = \frac{7}{4}$
​

Answer 1

Above the question is mentioned,this question's answer will be X=3.

Answer 2

$\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}$

$\\ \color{maroon} \bigstar \: \: \rm \log _{3} x+\log _{9} x+\log _{61} x=\frac{7}{4}$

$\[ \begin{array}{l} \displaystyle \rm\ \Rightarrow \frac{1}{\log _{x} 3}+\frac{1}{\log _{x} 9}+\frac{1}{\log _{x} 81}=\frac{7}{4} \\ \\ \displaystyle \rm\ \Rightarrow \frac{1}{\log _{x} 3^{1}}+\frac{1}{\log _{x} 3^{2}}+\frac{1}{\log _{x} 3^{4}}=\frac{7}{4} \\ \\ \displaystyle \rm\ \Rightarrow \frac{1}{\log _{x} 3}+\frac{1}{2 \log _{x} 3}+\frac{1}{4 \log _{x} 3}=\frac{7}{4} \\ \\ \displaystyle \rm\ \Rightarrow \frac{1}{\log _{x} 3}\left[1+\frac{1}{2}+\frac{1}{4}\right]=\frac{7}{4} \\ \\ \displaystyle \rm\ \Rightarrow \log _{x} 3 \times \frac{7}{4}=\frac{7}{4} \\ \\ \displaystyle \rm\Rightarrow \log _{x} 3=\frac{7}{4} \times \frac{4}{7}=1=\log _{3} 3 \qquad \qquad \{ \because \: \: log_{a}(a) = 1\} \end{array} \]$

Comparing, we get

$\color{green} \boxed{ \rm\[ \therefore x=3 \]}$