Answer: (a) \(\lg\frac{500a^2}{b}\); (b) \(c=\frac19\) or \(c=3^{\frac14}\).

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

(a) Use the laws of logarithms:

Also, since \(b\) is positive,

\(\frac12\lg(4b^2)=\lg\sqrt{4b^2}=\lg(2b).\)

Therefore

\(3+2\lg a-\frac12\lg(4b^2) =\lg1000+\lg a^2-\lg(2b).\)

So

\(3+2\lg a-\frac12\lg(4b^2) =\lg\frac{1000a^2}{2b} =\lg\frac{500a^2}{b}.\)

(b) Let

\(u=\log_c3.\)

Then

\(\log_3c=\frac1u.\)

The equation

\(2\log_c3=7+4\log_3c\)

becomes

\(2u=7+\frac4u.\)

Multiplying by \(u\),

\(2u^2=7u+4.\)

So

\(2u^2-7u-4=0.\)

Factorising,

\((2u+1)(u-4)=0.\)

Therefore

\(u=-\frac12 \quad\text{or}\quad u=4.\)

Since \(u=\log_c3\),

\(\log_c3=-\frac12 \quad\text{or}\quad \log_c3=4.\)

These give

\(c^{-\frac12}=3 \quad\text{or}\quad c^4=3.\)

Hence

\(c=\frac19 \quad\text{or}\quad c=3^{\frac14}.\)