Question

Solve for X


`\log _ { 3 } x = 9 \log _ { x } 3`
pls gv step by step explaination​

Answer 1

Answer: x=3 x=1/27

Explanation:

`log_3x=9log_x3\\`

The area of acceptable values:

`x > 0\ \ \ \ x\\eq 1\\Hence,\\x\in(0,1)U(1,+\infty)`

Solution:

`\displaystyle\\log_3x=(9)/(log_3x) \ \ \ \ \ \boxed{ log_ab=(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=((1)/(3))^3\\ x=(1)/(3)*(1)/(3)*(1)/(3) \\ x=(1)/(27)`

Answer 2

`x=27, \quad x= (1)/(27)`

Explanation:

`\boxed{\begin{minipage}{3.7 cm}\underline{Logs Laws}\\\\$\log_ba=(\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\\\\& =(9\log_33)/(\log_3x)\\\\&=(9(1))/(\log_3x)\\\\&=(9)/(\log_3x)\end{aligned}`

Therefore:

`\implies \log_3x=(9)/(\log_3x)`

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

`\implies u=(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 (1)/(3^3)=x`

`\implies x=(1)/(27)`