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 $\alpha$.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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

$\text{lo}{\text{g}}_{2}y+\text{lo}{\text{g}}_{4}x+\text{lo}{\text{g}}_{4}2x=0$

$\text{lo}{\text{g}}_{2}y+\text{lo}{\text{g}}_{4}2x^2=0$     M1

$\text{lo}{\text{g}}_{2}y+\frac{1}{2}\text{lo}{\text{g}}_{2}2x^2=0$     M1

$\text{lo}{\text{g}}_{2}y=-\frac{1}{2}\text{lo}{\text{g}}_{2}2x^2$

$\text{lo}{\text{g}}_{2}y=\text{lo}{\text{g}}_2\left(\frac{1}{\sqrt{2x}}\right)$     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 $\underset{1}{\overset{\alpha }{\int }}\frac{1}{\sqrt{2}}{x}^{-1}\text{d}x$     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]