Answer: (a) \(x=6.19\). (b) \(y=-0.281\).

(a) Take logarithms to base \(3\):

\(\frac{x}{2}-1=\log_3 10.\)

Therefore

\(\frac{x}{2}=1+\log_3 10, \qquad x=2(1+\log_3 10).\)

Using \(\log_3 10=\dfrac{\log 10}{\log 3}\),

\(x=2\left(1+\frac{\log 10}{\log 3}\right)=6.19\quad\text{to 3 significant figures}.\)

(b) Start with

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

Divide by \(3e^{1-2y}\):

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

So

\(e^{5y+1}=\frac{2}{3}.\)

Taking natural logarithms gives

\(5y+1=\ln\frac23.\)

Hence

\(y=\frac{\ln(2/3)-1}{5}=-0.281\quad\text{to 3 significant figures}.\)