Exponential and Logarithmic Functions: practice solutions, Grade 12 – download the PDF

1 Evaluating an exponential function ★★★

\( f(0)=3\cdot2^{0}=3\cdot1=3 \).

\( f(4)=3\cdot2^{4}=3\cdot16=48 \).

\( f(-1)=3\cdot2^{-1}=3\cdot\dfrac12=\dfrac32=1.5 \).

2 Growth or decay? ★★★

(a) \( b=1.08\gt1 \): growth of 8% per step.

(b) \( b=0.9\lt1 \): decay, \( 1-0.9=0.1 \), so 10% loss per step.

(c) \( b=1.5\gt1 \): growth of 50% per step.

(d) \( b=\tfrac13\lt1 \): decay; the quantity loses two thirds of its value, about 66.7%, per step.

3 From exponential to logarithmic form ★★★

Use \( b^{y}=x\iff\log_b x=y \).

(a) \( \log_2 128=7 \).

(b) \( \log 0.01=-2 \).

(c) \( \log_9 3=\dfrac12 \).

4 Logarithms without a calculator ★★★

(a) \( 3^{4}=81 \), so \( 4 \).

(b) \( 5^{-2}=\dfrac1{25} \), so \( -2 \).

(c) \( 4^{1/2}=2 \), so \( \dfrac12 \).

(d) \( \ln e^{5}=5 \) (inverse property).

(e) \( 10^{3}=1000 \), so \( 3 \).

5 Condensing logarithms ★★★

(a) \( \log_6(4\cdot9)=\log_6 36=2 \).

(b) \( \log_2\dfrac{40}{5}=\log_2 8=3 \).

(c) \( \log_3\dfrac{6^{2}}{4}=\log_3 9=2 \).

6 Matching bases ★★★

(a) \( 64=2^{6} \), so \( x=6 \).

(b) \( 125=5^{3} \), so \( x+1=3 \) and \( x=2 \).

(c) \( 0.001=10^{-3} \), so \( x=-3 \).

(d) \( 8=2^{3}=\left(\tfrac12\right)^{-3} \), so \( x=-3 \).

7 Town population ★★★

(a) The factor is \( 1+0.03=1.03 \), so \( P(t)=12000(1.03)^{t} \).

(b) \( P(2)=12000(1.0609)=12{,}730.8\approx 12{,}731 \).

The town has about 12,731 people after 2 years.

8 Simple logarithmic equations ★★★

(a) \( x=3^{4}=81 \).

(b) \( x-1=2^{3}=8 \), so \( x=9 \).

(c) \( x=e^{0}=1 \).

(d) \( x^{2}=49 \) and \( x\gt0 \), so \( x=7 \).

9 Using the change of base formula ★★★

(a) \( \log_5 40=\dfrac{\ln 40}{\ln 5}\approx\dfrac{3.68888}{1.60944}\approx 2.2920 \).

(b) \( \log_2 10=\dfrac{\ln 10}{\ln 2}\approx\dfrac{2.30259}{0.69315}\approx 3.3219 \).

10 An exponential equation with a calculator ★★★

Take \( \ln \) of both sides: \( x\ln 7=\ln 30 \).

Exact: \( x=\dfrac{\ln 30}{\ln 7} \). Approximate: \( \dfrac{3.40120}{1.94591}\approx 1.7479 \).

Check: \( 7^{1.7479}\approx 30 \). Good.

11 Equation with base e ★★★

Divide by 4: \( e^{2x}=9 \).

Take \( \ln \): \( 2x=\ln 9 \), so \( x=\dfrac{\ln 9}{2}=\ln 3\approx 1.0986 \).

12 A logarithmic equation ★★★

Exponential form: \( 3x+1=4^{2}=16 \), so \( 3x=15 \) and \( x=5 \).

Check: \( 3(5)+1=16\gt0 \) and \( \log_4 16=2 \). Correct.

13 Monthly compounding ★★★

Here \( P=2500 \), \( r=0.048 \), \( n=12 \), \( t=5 \), so \( \dfrac rn=0.004 \).

\( A=2500(1.004)^{60}\approx 2500(1.27064)\approx\$3{,}176.60 \).

14 When does the car lose enough value? ★★★

Solve \( 24000(0.85)^{t}=10000 \): \( 0.85^{t}=\dfrac{5}{12} \), so \( t=\dfrac{\ln(5/12)}{\ln0.85}\approx 5.39 \).

Check: \( V(5)\approx\$10{,}648.93 \) and \( V(6)\approx\$9{,}051.58 \).

The car is worth less than $10,000 during the 6th year; after 6 whole years it is first below that value.

15 Medicine in the body ★★★

(a) \( N(10)=80\cdot0.5^{10/6}\approx 80(0.31498)\approx 25.2\text{ mg} \).

(b) \( 10=80\cdot0.5^{t/6}\Rightarrow0.5^{t/6}=\dfrac18=0.5^{3}\Rightarrow\dfrac t6=3\Rightarrow t=18 \).

Ten milligrams remain after 18 hours (three half-lives), as the graph shows.

16 Expanding a logarithm ★★★

Quotient law: \( \ln(x^{3}\sqrt y)-\ln e^{2} \).

Product and power laws: \( 3\ln x+\tfrac12\ln y-2 \).

Answer: \( 3\ln x+\dfrac12\ln y-2 \).

17 Equation in quadratic form ★★★

Since \( 9^{x}=(3^{x})^{2} \), let \( u=3^{x}\gt0 \): \( u^{2}-4u-45=0 \Rightarrow(u-9)(u+5)=0 \).

\( u=-5 \) is impossible because \( 3^{x}\gt0 \). So \( 3^{x}=9 \) and \( x=2 \).

Check: \( 81-36-45=0 \).

18 Two logs, one extraneous root ★★★

Condense: \( \log[x(x-21)]=2\Rightarrow x(x-21)=10^{2}=100 \).

\( x^{2}-21x-100=0\Rightarrow(x-25)(x+4)=0 \).

Domain: \( x\gt21 \), so \( x=-4 \) is rejected. Answer: \( x=25 \).

Check: \( \log25+\log4=\log100=2 \).

19 Equating arguments ★★★

Condense the left side: \( \ln[(x+2)(x-2)]=\ln5\Rightarrow x^{2}-4=5\Rightarrow x^{2}=9 \).

So \( x=3 \) or \( x=-3 \). The domain requires \( x\gt2 \), so \( x=-3 \) is rejected.

Answer: \( x=3 \).

20 Doubling and tripling time ★★★

(a) \( Pe^{0.045t}=2P\Rightarrow t=\dfrac{\ln2}{0.045}\approx15.40 \) years.

(b) \( Pe^{0.06t}=3P\Rightarrow t=\dfrac{\ln3}{0.06}\approx18.31 \) years.

21 Chemistry: pH ★★★

(a) \( \text{pH}=-\log(3.2\times10^{-5})=-(\log3.2-5)\approx-(0.5051-5)\approx4.49 \).

(b) \( -\log[\text{H}^{+}]=7.4\Rightarrow[\text{H}^{+}]=10^{-7.4}\approx3.98\times10^{-8} \) mol/L.

22 Fitting a model to two data points ★★★

(a) \( 800e^{4k}=5000\Rightarrow e^{4k}=6.25\Rightarrow k=\dfrac{\ln6.25}{4}=\dfrac{\ln2.5}{2}\approx0.4581 \).

(b) \( N(6)=800e^{6k}=800(6.25)^{1.5}=800(15.625)=12{,}500 \) cells.

23 True or false? ★★★

(a) False. With \( a=b=10 \): \( \log20\approx1.301 \) but \( \log10+\log10=2 \).

(b) False. For \( x=-3 \), \( \ln9\approx2.197 \) but \( \ln(-3) \) does not exist. The correct identity is \( \ln(x^{2})=2\ln|x| \).

(c) True. \( \log_b x\cdot\log_x b=\dfrac{\ln x}{\ln b}\cdot\dfrac{\ln b}{\ln x}=1 \).

24 Annual versus continuous ★★★

(a) Solve \( 800(1.07)^{t}=5000\Rightarrow1.07^{t}=6.25\Rightarrow t=\dfrac{\ln6.25}{\ln1.07}\approx27.09 \). Check: \( 800(1.07)^{27}\approx\$4{,}971 \) and \( 800(1.07)^{28}\approx\$5{,}319 \). The balance first passes $5,000 at the end of year 28.

(b) \( 800e^{0.07t}=5000\Rightarrow t=\dfrac{\ln6.25}{0.07}\approx26.18 \) years.