Answer: (a) \(\displaystyle \lg\!\left(\frac{x^2}{3(x+6)}\right)\). (b) \(x=6\).

Answer: (a) \(\displaystyle \lg\!\left(\frac{x^2}{3(x+6)}\right)\). (b) \(x=6\).

(a) First use \(2\lg x=\lg x^2\), and combine the logarithms inside the brackets:

\(\lg(x+6)+\lg 3=\lg(3(x+6))\).

Therefore

\(2\lg x-\{\lg(x+6)+\lg3\}=\lg x^2-\lg(3(x+6))\).

Using \(\lg A-\lg B=\lg\!\left(\frac{A}{B}\right)\), this is

\(\displaystyle \lg\!\left(\frac{x^2}{3(x+6)}\right)\).

(b) The equation becomes

\(\displaystyle \lg\!\left(\frac{x^2}{3(x+6)}\right)=0\).

Since \(\lg 1=0\),

\(\displaystyle \frac{x^2}{3(x+6)}=1\).

Thus

\(x^2=3x+18\),

so

\(x^2-3x-18=0\).

Factorising gives

\((x-6)(x+3)=0\).

The logarithms require \(x\gt 0\), so \(x=-3\) is rejected. Hence

\(x=6\).