Question

Solve for X "TexFormula1" title="\log _ { 3 } x = 9 \log _ { x } 3" alt="\log _ { 3 } x = 9 \log _ { x } 3" align="absmiddle" class="latex-formula">
pls gv step by step explaination​

Answer 1

Answer:

$x=27, \quad x= \dfrac{1}{27}$

Step-by-step explanation:

$\boxed{\begin{minipage}{3.7 cm}\underline{Logs Laws}\\\\ \log_ba=\dfrac{\log_ca}{\log_cb} \\\\\\ \log_ab=c \iff a^c=b \\\end{minipage}}$

Given equation:

$\log_3x=9\log_x3$

Change the base of  9logₓ3  to base 3:

$\begin{aligned}\implies \log_3x & =9\log_x3\\\\& =\dfrac{9\log_33}{\log_3x}\\\\&=\dfrac{9(1)}{\log_3x}\\\\&=\dfrac{9}{\log_3x}\end{aligned}$

Therefore:

$\implies \log_3x=\dfrac{9}{\log_3x}$

$\textsf{Let }\log_3x=u:$

$\implies u=\dfrac{9}{u}$

$\implies u^2=9$

$\implies u=\pm3$

$\textsf{Substitute back in }u=\log_3x:$

$\implies \log_3x=\pm3$

Case 1

$\implies \log_3x=3$

$\implies 3^3=x$

$\implies x=27$

Case 2

$\implies \log_3x=-3$

$\implies 3^{-3}=x$

$\implies \dfrac{1}{3^3}=x$

$\implies x=\dfrac{1}{27}$

Answer 2

Answer: x=3   x=1/27

Step-by-step explanation:

$log_3x=9log_x3$

The area of acceptable values:

$x > 0\ \ \ \ x\neq 1\\Hence,\\x\in(0,1)U(1,+\infty)$

Solution:

$\displaystyle\\log_3x=\frac{9}{log_3x} \ \ \ \ \ \boxed{ log_ab=\frac{1}{log_ba} }\\\\Multiply\ both\ parts\ of\ the\ equation\ by\ log_3x :\\\\log^2_3x=9\\log^2_3x=3^2\\ log_3x=б3\\\\log_3x=3\\\\x=3^3\\x=3*3*3\\x=27\\\\\\log_3x=-3\\x=3^{-3}\\\displaystyle\\x=(\frac{1}{3})^3\\ x=\frac{1}{3}*\frac{1}{3}*\frac{1}{3} \\ x=\frac{1}{27}$