Answer: \(x=\ln\frac{23}{5}\), \(y=\ln\frac25\); \(t=\frac{2-\ln5}{3}\).

The key point is to reduce the equation to a standard trigonometric equation, then use the given interval to list all valid solutions.

(a) Let

\(X=e^x,\qquad Y=e^y.\)

The equations become

\(X+Y=5\)

and

\(2X-3Y=8.\)

Multiply the first equation by \(3\):

\(3X+3Y=15.\)

Add this to \(2X-3Y=8\):

\(5X=23.\)

So

\(X=\frac{23}{5}.\)

Then

\(Y=5-\frac{23}{5}=\frac25.\)

Therefore

\(e^x=\frac{23}{5},\qquad e^y=\frac25.\)

Taking natural logarithms,

\(x=\ln\frac{23}{5},\qquad y=\ln\frac25.\)

(b) From

\(e^{2t-1}=5e^{5t-3},\)

divide by \(e^{5t-3}\):

\(e^{(2t-1)-(5t-3)}=5.\)

Simplifying the exponent,

\(e^{2-3t}=5.\)

Take natural logarithms:

\(2-3t=\ln5.\)

Thus

\(t=\frac{2-\ln5}{3}.\)