Answer: (a) \(\displaystyle \lg\left(\frac{x^3}{1000y^4}\right)\); (b) \(x=\sqrt3\) or \(x=9\).

Use the laws of logarithms and exponentials first, then check that any logarithm arguments are positive where required.

(a) Use the laws of logarithms:

\(3\lg x=\lg x^3.\)

Also,

\(2\lg y^2=\lg (y^2)^2=\lg y^4.\)

Since \(3=\lg1000\),

\(3\lg x-2\lg y^2-3 =\lg x^3-\lg y^4-\lg1000.\)

Therefore

\(3\lg x-2\lg y^2-3 =\lg\left(\frac{x^3}{1000y^4}\right).\)

(b) Let

\(a=\log_3 x.\)

Then

\(\log_x3=\frac{1}{\log_3x}=\frac1a.\)

The equation becomes

\(a+\frac1a=\frac52.\)

Multiplying by \(2a\),

\(2a^2+2=5a.\)

So

\(2a^2-5a+2=0.\)

Factorising,

\((2a-1)(a-2)=0.\)

Hence

\(a=\frac12 \quad\text{or}\quad a=2.\)

So

\(\log_3x=\frac12 \quad\text{or}\quad \log_3x=2.\)

Therefore

\(x=\sqrt3 \quad\text{or}\quad x=9.\)