Answer: (a) \(x=-8.32\); (b) \(y=0.0493\).

Answer: (a) \(x=-8.32\); (b) \(y=0.0493\).

(a) Take logarithms of both sides:

\((3x+1)\log2=(x-2)\log5.\)

Expand and collect the \(x\)-terms:

\(3x\log2+\log2=x\log5-2\log5.\)

So

\(x(3\log2-\log5)=-\log2-2\log5.\)

Therefore

\(x=\frac{-\log2-2\log5}{3\log2-\log5}.\)

This gives

\(x=-8.32\quad\text{to 3 significant figures.}\)

(b) Let

\(u=e^{2y+1}.\)

The equation becomes

\(u=1+\frac6u.\)

Multiplying by \(u\),

\(u^2=u+6.\)

So

\(u^2-u-6=0.\)

Factorising,

\((u-3)(u+2)=0.\)

Since \(u=e^{2y+1}\gt 0\), we take \(u=3\). Hence

\(e^{2y+1}=3.\)

Taking natural logarithms,

\(2y+1=\ln3, \qquad y=\frac{\ln3-1}{2}.\)

Thus

\(y=0.0493\quad\text{to 3 significant figures.}\)