Exponential and Logarithmic Functions: practice questions

6 questions · 11 marks · free, no account · Read the notes first

1. Question 1Multiple choice · 1 mark

   $\log 2 + \log 50 =$

   A

   $\log 52$

   B

   $2$

   C

   $100$

   D

   $1$

2. Question 2Multiple choice · 1 mark

   If $2^{x + 1} = 32$, then $x =$

   A

   $5$

   B

   $16$

   C

   $4$

   D

   $3$

3. Question 3Multiple choice · 1 mark

   $\log_3 81 - \log_3 9 =$

   A

   $2$

   B

   $9$

   C

   $\log_3 72$

   D

   $3$

4. Question 4Multiple choice · 1 mark

   If $\log x = a$ and $\log y = b$, then $\log \dfrac{x^2}{\sqrt{y}} =$

   A

   $a^2 - \sqrt{b}$

   B

   $2a + \dfrac{b}{2}$

   C

   $\dfrac{4a}{b}$

   D

   $2a - \dfrac{b}{2}$

5. Question 53 marks

   Solve $3^x = 20$, giving your answer correct to 3 significant figures.

   Show worked solution

   $$x \log 3 = \log 20 \qquad \text{1M}$$ $$x = \frac{\log 20}{\log 3} \qquad \text{1M}$$ $$x \approx 2.73 \qquad \text{1A}$$
6. Question 64 marks

   The value of a car after $t$ years is $$V$, where $V = 240\,000(0.85)^t$.

   (a) Find the value of the car after 3 years.

   (b) After how many full years will the value of the car first fall below $100 000?

   Show worked solution

   (a) $V = 240\,000(0.85)^3 = 147\,390$. The value is $147 390. 1A

   (b) $240\,000(0.85)^t < 100\,000$ 1M

   $$t \log 0.85 < \log \frac{5}{12}$$

   Since $\log 0.85 < 0$, the inequality sign reverses:

   $$t > \frac{\log(5/12)}{\log 0.85} = 5.387\ldots \qquad \text{1M}$$

   So the value first falls below $100 000 after 6 years. 1A