Answer: (a) \(x=\frac23(\ln10-2)\); (b) \(y=3\) or \(y=9\).

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

(a) Use the laws of indices:

\(\frac{(\mathrm e^{x+1})^2}{\sqrt{\mathrm e^x}} =\frac{\mathrm e^{2x+2}}{\mathrm e^{x/2}}.\)

So the equation becomes

\(\mathrm e^{2x+2-x/2}=10.\)

Hence

\(\mathrm e^{\frac32x+2}=10.\)

Taking natural logarithms,

\(\frac32x+2=\ln10.\)

Therefore

\(x=\frac23(\ln10-2).\)

(b) Use \(2\log_9y=\log_9(y^2)\):

\(\log_9(y^2)-\log_9(4y-9)=\frac12.\)

So

\(\log_9\left(\frac{y^2}{4y-9}\right)=\frac12.\)

Since \(9^{1/2}=3\),

\(\frac{y^2}{4y-9}=3.\)

Thus

\(y^2=12y-27,\)

so

\(y^2-12y+27=0.\)

Factorising,

\((y-3)(y-9)=0.\)

Both values satisfy the logarithm domain conditions, so

\(y=3\quad\text{or}\quad y=9.\)

This completes the solution and gives the required result.