Answer: (a) \(\lg\left(\dfrac{10(x+1)}{x-1}\right)\); (b) \(x=-1+5\sqrt5,\ -1+\dfrac{\sqrt5}{25}\).

We start with the main method. Use logarithm laws and index laws first so that the equation is reduced to an algebraic equation.

For part (a), write \(1=\lg 10\).

Then

\(1+\lg(x^2-1)-2\lg(x-1)\)

\(=\lg 10+\lg(x^2-1)-\lg((x-1)^2).\)

Using the logarithm laws,

\(\lg 10+\lg(x^2-1)-\lg((x-1)^2) =\lg\left(\frac{10(x^2-1)}{(x-1)^2}\right).\)

Since \(x^2-1=(x-1)(x+1)\), and \(x\gt 1\), this simplifies to

\(\lg\left(\frac{10(x+1)}{x-1}\right).\)

For part (b), let

\(u=\log_5(x+1).\)

By change of base,

\(\log_{x+1}5=\frac{1}{\log_5(x+1)}=\frac1u.\)

The equation becomes

\(4u=\frac9u.\)

So

\(4u^2=9,\)

and

\(u=\pm\frac32.\)

Therefore

\(x+1=5^{3/2}\quad\text{or}\quad x+1=5^{-3/2}.\)

Hence

\(x=-1+5\sqrt5\quad\text{or}\quad x=-1+\frac{\sqrt5}{25}.\)

This completes the solution and gives the required result.