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]