Answer: (a) \(y=1.604\); (b) \(x=0\) or \(x=\frac12\ln3\).

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

(a) Take logarithms of both sides:

\((2y-1)\log5=\log6+y\log3.\)

Expand and collect the terms involving \(y\):

\(2y\log5-\log5=\log6+y\log3,\)

so

\(y(2\log5-\log3)=\log6+\log5.\)

Hence

\(y=\frac{\log30}{2\log5-\log3}.\)

Therefore

\(y=1.604\)

to 3 decimal places.

(b) Multiply the equation by \(e^{2x}\):

\(e^{4x}-4e^{2x}+3=0.\)

Let \(u=e^{2x}\). Then

\(u^2-4u+3=0.\)

Factorising,

\((u-1)(u-3)=0.\)

So

\(e^{2x}=1 \quad\text{or}\quad e^{2x}=3.\)

Therefore

\(x=0 \quad\text{or}\quad x=\frac12\ln3.\)