Sorry please help me.with another one, how should I do lg(x-2)=(lg3)^2. Thanks really much!
The notation $\displaystyle \lg$ is not familiar to me (I don't know for sure which base is implied), but I would write:
$\displaystyle \lg(x-2)=\lg(3)\lg(3)$
Now, on the right, use the property $\displaystyle c\cdot\log_a(b)=\log_a(b^c)$. What do you find?
You would only be able to use this property if the right hand side was $\displaystyle \displaystyle \begin{align*} \log{(3)} + \log{(3)} = 2\log{(3)} = \log{\left( 3^2 \right)} \end{align*}$, NOT $\displaystyle \displaystyle \begin{align*} \log{(3)} \log{(3)} = \left[ \log{(3)} \right]^2 \mathbf{\neq} \log{\left( 3^2 \right)} \end{align*}$.
To solve this equation...
$\displaystyle \displaystyle \begin{align*} \log{(x - 2)} &= \left[ \log{(3)} \right]^2 \\ x - 2 &= 10^{\left[ \log{(3)} \right]^2} \\ x &= 10^{\left[ \log{(3)} \right]^2} + 2 \end{align*}$