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Date	May 2019	Marks available	7	Reference code	19M.1.AHL.TZ2.H_7
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 2
Command term	Solve	Question number	H_7	Adapted from	N/A

Question

Solve the simultaneous equations

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

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

Markscheme

* 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

[7 marks]

Examiners report

[N/A]

Syllabus sections

Topic 1—Number and algebra » SL 1.5—Intro to logs

22M.1.SL.TZ2.8b:
Find the value of $k$ and the value of $p$ .
19N.1.SL.TZ0.S_10c:
The line ${L_2}$ is tangent to the graph of $g$ at ${\text{A}}$ and has equation $y = \left( {{\text{ln}}\,p} \right)x + q + 1$ .

The line ${L_2}$ passes through the point $\left( { - 2{\text{, }} - 2} \right)$ .

The gradient of the normal to $g$ at ${\text{A}}$ is $\frac{1}{{{\text{ln}}\left( {\frac{1}{3}} \right)}}$ .

Find the equation of ${L_1}$ in terms of $x$ .
19N.1.SL.TZ0.S_10a:
Write down the coordinates of ${\text{B}}$ .
18M.1.AHL.TZ1.H_5:
Solve ${\left( {{\text{ln}}\,x} \right)^2} - \left( {{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x} \right) < 2{\left( {{\text{ln}}\,2} \right)^2}$ .
17N.1.AHL.TZ0.H_1:
Solve the equation ${\log _2}(x + 3) + {\log _2}(x - 3) = 4$ .
18M.2.SL.TZ1.S_8b:
Use the regression equation to estimate the value of y when x = 3.57.
18M.1.AHL.TZ2.H_11a:
Show that ${\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{2}{\text{lo}}{{\text{g}}_r}\,x$ where $r,\,x \in {\mathbb{R}^ + }$ .
17M.2.AHL.TZ2.H_6:
Given that ${\log _{10}}\left( {\frac{1}{{2\sqrt 2 }}\left( {p + 2q} \right)} \right) = \frac{1}{2}\left( {{{\log }_{10}}p + {{\log }_{10}}q} \right),{\text{ }}p > 0,{\text{ }}q > 0$ , find $p$ in terms of $q$ .
18M.2.SL.TZ1.S_8a:
Find the value of a and of b.
19N.1.SL.TZ0.S_10b:
Given that $f'\left( a \right) = \frac{1}{{{\text{ln}}\,p}}$ , find the equation of ${L_1}$ in terms of $x$ , $p$ and $q$ .
17M.1.AHL.TZ1.H_1:
Find the solution of ${\log _2}x - {\log _2}5 = 2 + {\log _2}3$ .
16N.1.AHL.TZ0.H_7:
Solve the equation ${4^x} + {2^{x + 2}} = 3$ .
18M.2.SL.TZ1.S_8c:
The relationship between x and y can be modelled using the formula y = kxⁿ, where k ≠ 0 , n ≠ 0 , n ≠ 1.

By expressing ln y in terms of ln x, find the value of n and of k.

Topic 1—Number and algebra » SL 1.7—Laws of exponents and logs

Topic 1—Number and algebra

Question

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