Question

SUM OF ALL THE VALUES OF "X"?​

Answer 1

Given,

$\longrightarrow\sf{3^{\left(\log_3x\right)^2}+x^{\log_3x}=162}$

$\longrightarrow\sf{3^{\log_3x\,\cdot\,\log_3x}+x^{\log_3x}=162}$

Since $\sf{a^{mn}=(a^m)^n,}$

$\longrightarrow\sf{\left(3^{\log_3x}\right)^{\log_3x}+x^{\log_3x}=162}$

Since $\sf{a=b^{\log_ba},}$

$\longrightarrow\sf{x^{\log_3x}+x^{\log_3x}=162\quad\quad[\,\because\ 3^{\log_3x}=x\,]}$

$\longrightarrow\sf{2x^{\log_3x}=162}$

$\longrightarrow\sf{x^{\log_3x}=81}$

Taking log to the base $\sf{x,}$

$\longrightarrow\sf{\log_3x=\log_x81}$

Since $\sf{\log_ba=\dfrac{\log a}{\log b},}$

$\longrightarrow\sf{\dfrac{\log x}{\log3}=\dfrac{\log81}{\log x}}$

$\longrightarrow\sf{(\log x)^2=\log3\,\log81}$

$\longrightarrow\sf{(\log x)^2=\log3\,\log(3^4)}$

Since $\sf{\log(a^b)=b\log a,}$

$\longrightarrow\sf{(\log x)^2=\log3\cdot4\log3}$

$\longrightarrow\sf{(\log x)^2=4\left(\log3\right)^2}$

$\longrightarrow\sf{\log x=\pm\sqrt{4\left(\log3\right)^2}}$

$\longrightarrow\sf{\log x=\pm2\log3}$

$\longrightarrow\sf{\log x=2\log3\quad\ \ OR\quad\ \ \log x=-2\log3}$

$\longrightarrow\sf{\log x=\log(3^2)\quad\ \ OR\quad\ \ \log x=\log\left(3^{-2}\right)}$

$\longrightarrow\sf{\log x=\log9\quad\ \ OR\quad\ \ \log x=\log\left(\dfrac{1}{9}\right)}$

$\longrightarrow\sf{x=9\quad\ \ OR\quad\ \ x=\dfrac{1}{9}}$

So there are two possible values for $\sf{x.}$ Hence the sum of all values of $\sf{x}$ is,

$\longrightarrow\sf{S(x)=9+\dfrac{1}{9}}$

$\longrightarrow\sf{\underline{\underline{S(x)=\dfrac{82}{9}}}}$