Answer: (a) \(\lg\!\left(\frac{a^5}{1000b^4}\right)\). (b) \(x=4\) or \(x=-\frac45\).
(a) Use the logarithm laws
\(k\lg m=\lg(m^k),\qquad \lg M-\lg N=\lg\left(\dfrac MN\right).\)
Also, since the logarithms are base \(10\),
\(3=\lg1000.\)
Therefore
\(5\lg a-4\lg b-3=\lg a^5-\lg b^4-\lg1000.\)
Combining into a single logarithm gives
\(\lg\left(\dfrac{a^5}{1000b^4}\right).\)
(b) Let
\(u=\log_5(x+1).\)
Then
\(\log_{x+1}5=\dfrac1u.\)
The equation becomes
\(u-\dfrac1u=0.\)
Multiplying by \(u\),
\(u^2-1=0.\)
Thus
\(u=1\quad\text{or}\quad u=-1.\)
If \(u=1\), then
\(\log_5(x+1)=1,\)
so \(x+1=5\), giving \(x=4\).
If \(u=-1\), then
\(\log_5(x+1)=-1,\)
so \(x+1=5^{-1}=\dfrac15\), giving \(x=-\dfrac45\).
The logarithms require \(x+1\gt0\) and \(x+1\ne1\). Both values satisfy these conditions, so
\(x=4\quad\text{or}\quad x=-\dfrac45.\)
