1. ## SOLVE FOR x:

SOLVE FOR X: log ⁵√x - log1/93x =3

Any help will be much appreciated

2. ## Re: SOLVE FOR x:

What's the ᙮ meant to be?

3. ## Re: SOLVE FOR x:

Hello, Scoplex!

$\text{Solve for }x\!:\;\;\log_x\left(\sqrt[5]{x}\right) - \log_{\frac{1}{9}}\left(3^x\right) \:=\:3\;\;[1]$

$\text{Note that: }\:\log_x\left(\sqrt[5]{x}\right) \:=\:\log_x\left(x^{\frac{1}{5}}\right) \:=\:\tfrac{1}{5}\underbrace{\log_x(x)}_{\text{Thi s is 1}} \:=\:\tfrac{1}{5} \;\;[2]$

$\text{Let: }\,\log_{\frac{1}{9}}\left(3^x\right) \:=\:P\;\;[3]$

$\text{Then: }\:\left(\frac{1}{9}\right)^P \:=\:3^x \quad\Rightarrow\qiad \left(\frac{1}{3^2}\right)^P \:=\:3^x \quad\Rightarrow\quad \left(3^{-2}\right)^P \:=\:3^x$

. . . . . . $3^{-2P} \:=\:3^x \quad\Rightarrow\quad -2P \:=\:x \quad\Rightarrow\quad P \:=\:-\frac{x}{2}\;\;[4]$

$\text{Equate [3] and [4]: }\:\log_{\frac{1}{9}}\left(3^x\right) \;=\;-\frac{x}{2}\;\;[5]$

$\text{Substitute [2] and [5] into [1]: }\;\frac{1}{5} - \left(-\frac{x}{2}\right) \:=\:3$

$\text{Multiply by 10: }\;2 + 5x \:=\:30 \quad\Rightarrow\quad 5x \:=\:28 \quad\Rightarrow\quad \boxed{x \:=\:\frac{28}{5}}$