Logarithm equation

[123]

Hello guys, I need help to solve this:

\(\displaystyle \mathit{lg^2(100x)+lg^2(10x)=14+lg\frac{1}{x}}\)

what could be first action?

 

[galactus]

What is that superscript 2?. Is it the base or does it mean a square?. Just making sure.

 

[soroban]

Hello, 123!

\(\displaystyle [\log(100x)]^2 + [\log(10x)]^2\:=\:14+\log\left(\tfrac{1}{x}\right)\)

\(\displaystyle \text{We have: }\;\left[\log100+ \log x]^2 + [\log10 + \log x]^2 \;=\;14 + \log\left(x^{-1}\right)\)

. . . . . . . . . . . . . . \(\displaystyle [2 + \log x]^2 + [1 + \log x]^2 \;=\;14 - \log x\)

. \(\displaystyle 4 + 4\log x + (\log x)^2 + 1 + 2\log x + (\log x)^2 \;=\;14 - \log x\)

. . . . . . . . . . . . . . .\(\displaystyle 2(\log x)^2 + 7(\log x) - 9 \;=\;0\)

. . . . . . . . . . . . . . .\(\displaystyle (\log x - 1)(2\log x + 9) \;=\;0\)


\(\displaystyle \text{Therefore: }\;\begin{Bmatrix}\log x - 1 \:=\:0 & \Rightarrow & \log x \:=\:1 & \Rightarrow & x \:=\:10 \\ \\[-3mm] 2\log x - 9 \:=\:0 & \Rightarrow & \log x \:=\:\text{-}\frac{9}{2} & \Rightarrow & x \:=\:10\:\!^{\text{-}\frac{9}{2} \end{Bmatrix}\)