IB Questionbank

User interface language: English | Español

Question

Solve the simultaneous equations

$lo g_{2} 6 x = 1 + 2 lo g_{2} y$ ${\text{lo}}{{\text{g}}_2}6x = 1 + 2\,{\text{lo}}{{\text{g}}_2}y$

$1 + lo g_{6} x = lo g_{6} (15 y - 25)$ $1 + {\text{lo}}{{\text{g}}_6}x = {\text{lo}}{{\text{g}}_6}\left( {15y - 25} \right)$ .

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

use of at least one “log rule” applied correctly for the first equation M1

$lo g_{2} 6 x = lo g_{2} 2 + 2 lo g_{2} y$ ${\text{lo}}{{\text{g}}_2}6x = {\text{lo}}{{\text{g}}_2}2 + 2\,{\text{lo}}{{\text{g}}_2}y$

$= lo g_{2} 2 + lo g_{2} y^{2}$ $= {\text{lo}}{{\text{g}}_2}2 + \,{\text{lo}}{{\text{g}}_2}{y^2}$

$= lo g_{2} (2 y^{2})$ $= {\text{lo}}{{\text{g}}_2}\left( {2{y^2}} \right)$

$\Rightarrow 6 x = 2 y^{2}$ $\Rightarrow 6x = 2{y^2}$ A1

use of at least one “log rule” applied correctly for the second equation M1

$lo g_{6} (15 y - 25) = 1 + lo g_{6} x$ ${\text{lo}}{{\text{g}}_6}\left( {15y - 25} \right) = 1 + {\text{lo}}{{\text{g}}_6}x$

$= lo g_{6} 6 + lo g_{6} x$ $= {\text{lo}}{{\text{g}}_6}6 + {\text{lo}}{{\text{g}}_6}x$

$= lo g_{6} 6 x$ $= {\text{lo}}{{\text{g}}_6}6x$

$\Rightarrow 15 y - 25 = 6 x$ $\Rightarrow 15y - 25 = 6x$ A1

attempt to eliminate $x$ $x$ (or $y$ $y$ ) from their two equations M1

$2 y^{2} = 15 y - 25$ $2{y^2} = 15y - 25$

$2 y^{2} - 15 y + 25 = 0$ $2{y^2} - 15y + 25 = 0$

$(2 y - 5) (y - 5) = 0$ $\left( {2y - 5} \right)\left( {y - 5} \right) = 0$

$x = \frac{25}{12}, y = \frac{5}{2},$ $x = \frac{{25}}{{12}}{\text{,}}\,\,y = \frac{5}{2}{\text{,}}$ A1

or $x = \frac{25}{3}, y = 5$ $x = \frac{{25}}{3}{\text{,}}\,\,y = 5$ A1

Note: $x$ $x$ , $y$ $y$ values do not have to be “paired” to gain either of the final two A marks.

[7 marks]

[N/A]