log with a base of x(x)^1/5-log with a base of fracion1/9 3^x=3.i have no idea.thanks
P poplap New member Joined Mar 4, 2012 Messages 2 Mar 4, 2012 #1 log with a base of x(x)^1/5-log with a base of fracion1/9 3^x=3.i have no idea.thanks
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Mar 4, 2012 #2 Hello, poplap! What an ugly problem . . . \(\displaystyle \log_x(x^{\frac{1}{5}}) - \log_{\frac{1}{9}}\left(3^x\right) \:=\:3\) Click to expand... Note that: .\(\displaystyle \log_x(x^{\frac{1}{5}}) \:=\:\frac{1}{5}\) \(\displaystyle \text{Let: }\,\log_{\frac{1}{9}}(3^x) \:=\ \quad\Rightarrow\quad \left(\dfrac{1}{9}\right)^P \:=\:3^x \quad\Rightarrow\quad \left(\dfrac{1}{3^2}\right)^P \:=\:3^x \) . . . \(\displaystyle \left(3^{\text{-}2}\right)^P \:=\:3^x \quad\Rightarrow\quad 3^{\text{-}2P} \:=\:3^x \quad\Rightarrow\quad \text{-}2P \:=\:x \quad\Rightarrow\quad P \:=\:\text{-}\frac{x}{2}\) Hence: .\(\displaystyle \log_{\frac{1}{9}}(3^x) \:=\:\text{-}\frac{x}{2}\) The equation becomes: .\(\displaystyle \dfrac{1}{5} - \left(\text{-}\dfrac{x}{2}\right) \:=\:3 \) Multiply by 10: ..\(\displaystyle 2 + 5x \:=\:30 \quad\Rightarrow\quad 5x \:=\:28\) . . Therefore: .\(\displaystyle \boxed{x \;=\;\frac{28}{5}}\)
Hello, poplap! What an ugly problem . . . \(\displaystyle \log_x(x^{\frac{1}{5}}) - \log_{\frac{1}{9}}\left(3^x\right) \:=\:3\) Click to expand... Note that: .\(\displaystyle \log_x(x^{\frac{1}{5}}) \:=\:\frac{1}{5}\) \(\displaystyle \text{Let: }\,\log_{\frac{1}{9}}(3^x) \:=\ \quad\Rightarrow\quad \left(\dfrac{1}{9}\right)^P \:=\:3^x \quad\Rightarrow\quad \left(\dfrac{1}{3^2}\right)^P \:=\:3^x \) . . . \(\displaystyle \left(3^{\text{-}2}\right)^P \:=\:3^x \quad\Rightarrow\quad 3^{\text{-}2P} \:=\:3^x \quad\Rightarrow\quad \text{-}2P \:=\:x \quad\Rightarrow\quad P \:=\:\text{-}\frac{x}{2}\) Hence: .\(\displaystyle \log_{\frac{1}{9}}(3^x) \:=\:\text{-}\frac{x}{2}\) The equation becomes: .\(\displaystyle \dfrac{1}{5} - \left(\text{-}\dfrac{x}{2}\right) \:=\:3 \) Multiply by 10: ..\(\displaystyle 2 + 5x \:=\:30 \quad\Rightarrow\quad 5x \:=\:28\) . . Therefore: .\(\displaystyle \boxed{x \;=\;\frac{28}{5}}\)