Without using mathematical tables or calculator, evaluate \(\frac{\frac{3}{2} \log 27-3 \log 5 \sqrt{5}}{\log 0.6}\)

Question

(a) Without using mathematical tables or calculator, evaluate \(\frac{\frac{3}{2} \log 27-3 \log 5 \sqrt{5}}{\log 0.6}\)

(b) Two linear transformations \(\mathrm{A}\) and \(\mathrm{B}\) in the \(O_{x y}\) plane, are defined by :
\[
\begin{aligned}
&A:(x, y)(x+2 y,-x+y) \\
&B:(x, y)(2 x+3 y, x+2 y)
\end{aligned}
\]
(i) Write down the matrices \(A\) and \(B\); (ii) Find the image of the point \(P(-2,2)\) under the linear transformation \(A\) followed by B.

Answer 1

(a) \(\frac{\frac{3}{2} \log 27-3 \log 5 \sqrt{5}}{\log 0.6}\)
\[
\begin{aligned}
&=\frac{\frac{3}{2} \log 3^{3}-3 \log 5^{\frac{3}{2}}}{\log \left(\frac{3}{5}\right)} \\
&=\frac{\frac{2}{3} \times 3 \log ^{3}-3 \times \frac{3}{2} \log 5}{\log 3-\log 5} \\
&=\frac{\frac{9}{2}(\log 3-\log 5)}{\log 3-\log 5} \\
&=\frac{9}{2}=4 \frac{1}{2}
\end{aligned}
\]
(b)(i) \(A=\left(\begin{array}{cc}1 & 2 \\ -1 & 1\end{array}\right), B=\left(\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right)\)
(ii) Transformation of A followed by \(B\) is \(B A\).
\[
\begin{aligned}
&B A=\left(\begin{array}{cc}
2 & 3 \\
1 & 2
\end{array}\right)\left(\begin{array}{cc}
1 & 2 \\
-1 & 1
\end{array}\right) \\
&=\left(\begin{array}{ll}
-1 & 7 \\
-1 & 4
\end{array}\right)
\end{aligned}
\]
Image of \((-2,2)\) is given by

\(\left(\begin{array}{ll}-1 & 7 \\ -1 & 4\end{array}\right)\left(\begin{array}{c}-2 \\ 2\end{array}\right)\)
\(=\left(\begin{array}{l}16 \\ 10\end{array}\right)\)