Solve for x
- Thread starter poplap
- Start date
Hello, poplap!
What an ugly problem . . .
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: .\(\displaystyle \boxed{x \;=\;\frac{28}{5}}\)
What an ugly problem . . .
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: .\(\displaystyle \boxed{x \;=\;\frac{28}{5}}\)