Show that \({\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{2}{\text{lo}}{{\text{g}}_r}\,x\) where \(r,\,x \in {\mathbb{R}^ + }\).

Express \(y\) in terms of \(x\). Give your answer in the form \(y = p{x^q}\), where p , q are constants.

The region R, is bounded by the graph of the function found in part (b), the x-axis, and the lines \(x = 1\) and \(x = \alpha \) where \(\alpha  > 1\). The area of R is \(\sqrt 2 \).

Find the value of \(\alpha \).

METHOD 1

\({\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{{{\text{lo}}{{\text{g}}_r}\,{r^2}}}\left( { = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{{{\text{2}}\,{\text{lo}}{{\text{g}}_r}\,r}}} \right)\)     M1A1

\( = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{2}\)     AG

[2 marks]

METHOD 2

\({\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{{{\text{lo}}{{\text{g}}_x}\,{r^2}}}\)     M1

\( = \frac{1}{{2\,{\text{lo}}{{\text{g}}_x}\,r}}\)     A1

\( = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{2}\)     AG

[2 marks]

METHOD 1

\({\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0\)

\({\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,2{x^2} = 0\)     M1

\({\text{lo}}{{\text{g}}_2}\,y + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2{x^2} = 0\)     M1

\({\text{lo}}{{\text{g}}_2}\,y =  - \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2{x^2}\)

\({\text{lo}}{{\text{g}}_2}\,y = {\text{lo}}{{\text{g}}_2}\left( {\frac{1}{{\sqrt {2x} }}} \right)\)     M1A1

\(y = \frac{1}{{\sqrt 2 }}{x^{ - 1}}\)     A1

Note: For the final A mark, \(y\) must be expressed in the form \(p{x^q}\).

[5 marks]

METHOD 2

\({\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0\)

\({\text{lo}}{{\text{g}}_2}\,y + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,x + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2x = 0\)     M1

\({\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_2}\,{x^{\frac{1}{2}}} + {\text{lo}}{{\text{g}}_2}\,{\left( {2x} \right)^{\frac{1}{2}}} = 0\)     M1

\({\text{lo}}{{\text{g}}_2}\,\left( {\sqrt 2 xy} \right) = 0\)     M1

\(\sqrt 2 xy = 1\)     A1

\(y = \frac{1}{{\sqrt 2 }}{x^{ - 1}}\)     A1

Note: For the final A mark, \(y\) must be expressed in the form \(p{x^q}\).

[5 marks]

the area of R is \(\int\limits_1^\alpha  {\frac{1}{{\sqrt 2 }}} {x^{ - 1}}{\text{d}}x\)     M1

\( = \left[ {\frac{1}{{\sqrt 2 }}{\text{ln}}\,x} \right]_1^\alpha \)     A1

\( = \frac{1}{{\sqrt 2 }}{\text{ln}}\,\alpha \)     A1

\(\frac{1}{{\sqrt 2 }}{\text{ln}}\,\alpha  = \sqrt 2 \)     M1

\(\alpha  = {{\text{e}}^2}\)     A1

Note: Only follow through from part (b) if \(y\) is in the form \(y = p{x^q}\)

[5 marks]