Answer: (a) \(\displaystyle x=\frac13\ln\frac52\); (b) \((x,y)=\left(-14,\frac97\right)\) or \((9,-2)\).

Answer: (a) \(\displaystyle x=\frac13\ln\frac52\); (b) \((x,y)=\left(-14,\frac97\right)\) or \((9,-2)\).

(a) Let \(u=\mathrm e^{3x}\). Then \(\mathrm e^{6x}=u^2\), so the equation becomes

\(2u^2-3u-5=0.\)

Factorise:

\((2u-5)(u+1)=0.\)

Since \(u=\mathrm e^{3x}\gt 0\), reject \(u=-1\). Hence

\(\mathrm e^{3x}=\frac52.\)

Taking natural logarithms,

\(3x=\ln\frac52,\)

so

\(x=\frac13\ln\frac52.\)

(b) From the exponential equation,

\(\mathrm e^{4x-7-(5x+7y)}=\mathrm e^{-2}.\)

Therefore

\(4x-7-5x-7y=-2,\)

which simplifies to

\(x+7y=-5.\)

Also, \(xy+18=0\), so \(xy=-18\).

Using \(x=-5-7y\),

\(y(-5-7y)=-18.\)

So

\(7y^2+5y-18=0.\)

Factorising gives

\((7y-9)(y+2)=0.\)

Thus

\(y=\frac97\quad\text{or}\quad y=-2.\)

If \(y=\frac97\), then \(x=-14\). If \(y=-2\), then \(x=9\). Hence

\((x,y)=\left(-14,\frac97\right)\quad\text{or}\quad(9,-2).\)