WrongNumberX
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Solve for X
[tex]\log _ { 3 } x = 9 \log _ { x } 3[/tex]
pls gv step by step explaination​

Sagot :

semsee45

Answer:

[tex]x=27, \quad x= \dfrac{1}{27}[/tex]

Step-by-step explanation:

[tex]\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}}[/tex]

Given equation:

[tex]\log_3x=9\log_x3[/tex]

Change the base of  9logₓ3  to base 3:

[tex]\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}[/tex]

Therefore:

[tex]\implies \log_3x=\dfrac{9}{\log_3x}[/tex]

[tex]\textsf{Let }\log_3x=u:[/tex]

[tex]\implies u=\dfrac{9}{u}[/tex]

[tex]\implies u^2=9[/tex]

[tex]\implies u=\pm3[/tex]

[tex]\textsf{Substitute back in }u=\log_3x:[/tex]

[tex]\implies \log_3x=\pm3[/tex]

Case 1

[tex]\implies \log_3x=3[/tex]

[tex]\implies 3^3=x[/tex]

[tex]\implies x=27[/tex]

Case 2

[tex]\implies \log_3x=-3[/tex]

[tex]\implies 3^{-3}=x[/tex]

[tex]\implies \dfrac{1}{3^3}=x[/tex]

[tex]\implies x=\dfrac{1}{27}[/tex]

Answer: x=3   x=1/27

Step-by-step explanation:

[tex]log_3x=9log_x3\\[/tex]

The area of acceptable values:

[tex]x > 0\ \ \ \ x\neq 1\\Hence,\\x\in(0,1)U(1,+\infty)[/tex]

Solution:

[tex]\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}[/tex]

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