Answer: (a)(ii) \(x=15\), \(\displaystyle y=\sqrt{\frac23}=\frac{\sqrt6}{3}\); (b) \(x=a\) or \(x=a^3\).

Answer: (a)(ii) \(x=15\), \(\displaystyle y=\sqrt{\frac23}=\frac{\sqrt6}{3}\); (b) \(x=a\) or \(x=a^3\).

(a)(i) From

\(\lg x+2\lg y=1,\)

we get

\(\lg(xy^2)=1.\)

Since \(1=\lg10\),

\(xy^2=10.\)

From \(x-3y^2=13\),

\(y^2=\frac{x-13}{3}.\)

Substitute this into \(xy^2=10\):

\(x\left(\frac{x-13}{3}\right)=10.\)

Therefore

\(x^2-13x=30,\)

so

\(x^2-13x-30=0.\)

(a)(ii) Factorise:

\(x^2-13x-30=(x-15)(x+2).\)

Since \(\lg x\) is defined only for \(x\gt0\), reject \(x=-2\). Hence \(x=15\).

Then

\(y^2=\frac{15-13}{3}=\frac23.\)

Since \(\lg y\) is defined only for \(y\gt0\),

\(y=\sqrt{\frac23}=\frac{\sqrt6}{3}.\)

(b) Let

\(u=\log_a x.\)

Then

\(\log_x a=\frac1u.\)

The equation becomes

\(u+\frac{3}{u}=4.\)

Multiplying by \(u\),

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

So

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

and \(u=1\) or \(u=3\).

Therefore

\(\log_a x=1\quad\text{or}\quad\log_a x=3.\)

Hence

\(x=a\quad\text{or}\quad x=a^3.\)