Answer: \(x=3\).

Answer: \(x=3\).

Use the laws of logarithms:

\(\lg(2x-1)+\lg(x+2)=\lg((2x-1)(x+2)).\)

Also, since \(2=\lg 100\),

\(2-\lg4=\lg100-\lg4=\lg25.\)

Therefore

\(\lg((2x-1)(x+2))=\lg25.\)

So

\((2x-1)(x+2)=25.\)

Expanding gives

\(2x^2+3x-2=25,\)

so

\(2x^2+3x-27=0.\)

Factorise:

\((2x+9)(x-3)=0.\)

The possible roots are \(x=-\frac92\) and \(x=3\). Since the logarithms require \(2x-1\gt 0\) and \(x+2\gt 0\), only \(x=3\) is valid.