18M.1.AHL.TZ2.H_11

pestleMathematicsAAHLPaper 118M· sl-1-5-intro-to-logssource ↗

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

Markscheme / solution

* 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]

Examiners’ report

[N/A]