IB Maths AA SL 1.7 Rational exponents and logarithm laws

[Maximum number: 10]

(a)

Calculate the value of each of the following logarithms:

[ 6 ]

(i)

Calculate the value of each of the following logarithms:

$\quad \log _{9} 3$ ;

[ 1 ]

valid approach to find the required logarithm

\begin{aligned} & 9^{x}=3 \text { OR } 3^{2 x}=3 \text { OR } 3=9^{\frac{1}{2}} \text { OR } \frac{\log _{3} 3}{\log _{3} 9} \\ & \log _{9} 3=\frac{1}{2} \end{aligned}

(ii)

Calculate the value of each of the following logarithms:

$\quad \log _{\sqrt{3}} 81$ .

[ 5 ]

$(\sqrt{3})^{x}=81$ OR $\frac{\log _{3} 81}{\log _{3} \sqrt{3}}$
(3) $\frac{x}{2}=3^{4}$ OR $\frac{x}{2}=4$ OR $\frac{4}{\frac{1}{2}}$
x=8

(b)

It is given that $\log _{a b} a=3$ , where $a, b \in \mathbb{R}^{+}, a b \neq 1$ .

[ 4 ]

(i)

It is given that $\log _{a b} a=3$ , where $a, b \in \mathbb{R}^{+}, a b \neq 1$ .

Show that $\log _{a b} b=-2$ .

[ 4 ]

Note: There are many valid approaches to the question, and the steps may be seen in different ways. Some possible methods are given here, but candidates may use a combination of one or more of these methods.

In all methods, the final A mark is awarded for working which leads directly to the AG.

METHOD 1

(a b)^{3}=a

attempt to isolate b or a power of b
correct working

\begin{aligned} & b=\frac{a}{a^{3} b^{2}} \text { OR } b^{3}=a^{-2} \text { OR } b^{-1}=(a b)^{2} \text { OR } b^{3}=\frac{1}{a^{2}} \\ & b=\frac{1}{a^{2} b^{2}} \text { OR } b=(a b)^{-2} \text { OR } 3 \log _{a b} b=-2 \log _{a b} a \text { OR }-\log _{a b} b=2 \log _{a b} a b \\ & \log _{a b} b=-2 \end{aligned}

METHOD 2

(a b)^{3}=a

taking logarithm to base a b on both sides

\log _{a b}(a b)^{3}=\log _{a b} a \quad \text { OR } \quad \log _{a b} a^{3} b^{3}=\log _{a b} a

correct application of log rules leading to equation in terms of $\log _{a b}$

\begin{aligned} & 3 \log _{a b} a+3 \log _{a b} b=\log _{a b} a \quad \text { OR } \quad 3 \log _{a b} b=-2 \log _{a b} a \quad \text { OR } \log _{a b} b^{3}=\log _{a b} a^{-2} \\ & \log _{a b} b=\log _{a b} a^{-\frac{2}{3}} \quad \text { OR } \quad \log _{a b} b=-\frac{2}{3} \log _{a b} a \quad \text { OR } \quad \log _{a b} b=-\frac{2}{3}(3) \\ & \log _{a b} b=-2 \end{aligned}

Note: Candidates may substitute $\log _{a b} a=3$ at any point in their working.

METHOD 3

\log _{a b} a=3

writing in terms of base a

\frac{\log _{a} a}{\log _{a} a b}(=3)

correct application of log rules

\begin{aligned} & \frac{\log _{a} a}{\log _{a} a+\log _{a} b}(=3) \quad \mathrm{OR} \frac{1}{1+\log _{a} b}(=3) \quad \mathrm{OR} 3 \log _{a} b=-2 \quad \mathrm{OR} \\ & \log _{a} b=-\frac{2}{3} \end{aligned}

writing $\log _{a b} b$ in terms of base a

\log _{a b} b=\frac{\log _{a} b}{\log _{a} a+\log _{a} b}

correct working

\log _{a b} b=\frac{-\frac{2}{3}}{1-\frac{2}{3}} \text { OR } \frac{\left(-\frac{2}{3}\right)}{\left(\frac{1}{3}\right)}

\log _{a b} b=-2

METHOD 4

\begin{aligned} & \log _{a b} a b=1 \\ & \log _{a b} a+\log _{a b} b=1 \\ & 3+\log _{a b} b=1 \\ & \log _{a b} b=-2 \end{aligned}

IB Maths AA SL 1.7 Rational Exponents and Logarithm Laws Questions

Exam points

IB Maths AA SL 1.7 Rational Exponents and Logarithm Laws Questions question 1

Question (a)

Question (i)

Question (ii)

Question (b)

Question (i)