Exponential Functions: practice solutions, Grade 11 – download the PDF

1 Evaluating an exponential function ★★★

\( f(0) = 4 \cdot 3^{0} = 4 \cdot 1 = 4 \).

\( f(1) = 4 \cdot 3 = 12 \).

\( f(2) = 4 \cdot 9 = 36 \).

\( f(-1) = 4 \cdot 3^{-1} = \dfrac{4}{3} \).

2 Growth or decay? ★★★

1. \( b = 1.08 > 1 \): growth of 8% per step.
2. \( b = 0.9 \), between 0 and 1: decay of 10% per step (\( 1 - 0.9 = 0.1 \)).
3. \( b = 1.5 > 1 \): growth of 50% per step.
4. \( b = 0.25 \): decay of 75% per step (\( 1 - 0.25 = 0.75 \)).

3 True or false? ★★★

1. True: the variable is in the exponent and the base \( 5 \) is positive and not 1.
2. False: the variable is the base, so this is a power function.
3. False: \( 1^{x} = 1 \) for every \( x \), so it is a constant function; exponential bases must differ from 1.
4. True: \( y = 2 \cdot 3^{0} = 2 \).
5. True: the base \( 0.5 \) is between 0 and 1, so the function decays.

4 Completing a table ★★★

The outputs are multiplied by \( 10 \div 5 = 2 \) each step, so \( b = 2 \) and \( f(3) = 20 \cdot 2 = 40 \).

Since \( f(0) = 5 \), \( a = 5 \) and \( f(x) = 5 \cdot 2^{x} \).

\( f(5) = 5 \cdot 2^{5} = 5 \cdot 32 = 160 \).

5 A growing town ★★★

(a) Growth of 3% gives \( b = 1.03 \), so \( P(t) = 12000(1.03)^{t} \).

(b) \( P(2) = 12000 \cdot 1.03^{2} = 12000 \cdot 1.0609 = 12730.8 \). After 2 years the town has about 12,731 people.

6 Percent to factor ★★★

(a) \( b = 1 + 0.06 = 1.06 \).

(b) \( b = 1 - 0.12 = 0.88 \).

(c) \( b = 1 + 0.025 = 1.025 \).

(d) \( b = 1 - 0.40 = 0.60 \).

7 Reading a graph ★★★

(a) The curve crosses the y-axis at \( A \), so the y-intercept is \( (0, 3) \).

(b) Point \( C \) gives \( h(2) = 12 \).

(c) The curve flattens toward the x-axis: \( y = 0 \).

(d) Point \( D \) gives \( h(3) = 24 \), so \( x = 3 \).

(e) The outputs double each step, so \( h(x) = 3 \cdot 2^{x} \) is a growth function.

8 A phone that loses value ★★★

(a) \( b = 1 - 0.20 = 0.8 \), so \( V(t) = 900(0.8)^{t} \).

(b) \( V(3) = 900 \cdot 0.512 = 460.80 \): about 460.80 dollars.

(c) \( V(4) = 900 \cdot 0.4096 = 368.64 < 400 \), while \( V(3) = 460.80 > 400 \). The phone is worth less than 400 dollars after 4 years.

9 Equation from two points ★★★

\( f(0) = a = 3 \).

\( f(2) = 3b^{2} = 48 \), so \( b^{2} = 16 \). Since \( b > 0 \), \( b = 4 \).

The formula is \( f(x) = 3 \cdot 4^{x} \), and \( f(3) = 3 \cdot 64 = 192 \).

10 Monthly or yearly? ★★★

(a) \( A = 1500\left(1 + \dfrac{0.05}{12}\right)^{36} \approx 1742.21 \) dollars.

(b) \( A = 1500(1.05)^{3} = 1500 \cdot 1.157625 = 1736.4375 \approx 1736.44 \) dollars.

Monthly compounding earns \( 1742.21 - 1736.44 = 5.77 \) dollars more.

11 Shifting left and down ★★★

(a) Shift the graph of \( y = 3^{x} \) 2 units left and 4 units down.

(b) The asymptote moves from \( y = 0 \) to \( y = -4 \); the range is \( y > -4 \).

(c) \( g(0) = 3^{2} - 4 = 5 \) and \( g(-2) = 3^{0} - 4 = -3 \).

12 A reflected exponential ★★★

(a) Reflect \( y = 2^{x} \) across the x-axis, then shift it up 5 units.

(b) The asymptote is \( y = 5 \). The curve lies below it, so the range is \( y < 5 \).

(c) \( h(0) = -1 + 5 = 4 \) and \( h(3) = -8 + 5 = -3 \).

13 Doubling every three hours ★★★

(a) \( N(9) = 400 \cdot 2^{3} = 3200 \) cells.

(b) \( N(12) = 400 \cdot 2^{4} = 6400 \) cells.

(c) \( N(1) = 400 \cdot 2^{1/3} \approx 400 \cdot 1.26 \approx 504 \) cells.

14 Find the mistakes ★★★

(a) Maya multiplied \( 2 \) and \( 3 \) before applying the exponent. By the order of operations, the exponent comes first: \( 2 \cdot 3^{2} = 2 \cdot 9 = 18 \).

(b) Leo looked at \( a = 0.4 \), but growth or decay depends on the base \( b = 2 \). Since \( b > 1 \), the function is a growth function that starts at 0.4.

15 Which grows faster? ★★★

Values: \( A \): 200, 300, 450, 675, 1012.5, 1518.75, 2278.125. \( B \): 500, 600, 720, 864, 1036.8, 1244.16, 1492.992.

At \( x = 4 \), \( 1012.5 < 1036.8 \); at \( x = 5 \), \( 1518.75 > 1244.16 \). So \( A \) first exceeds \( B \) at \( x = 5 \) years.

The base 1.5 is larger than 1.2, so \( A \) is multiplied by a bigger factor each year; a larger base always beats a larger starting value in the long run.

16 Half-life ★★★

(a) \( A(18) = 80 \cdot \left(\tfrac{1}{2}\right)^{3} = 10 \) grams.

(b) \( A(9) = 80 \cdot 2^{-1.5} \approx 80 \cdot 0.35355 \approx 28.3 \) grams.

(c) Halving repeatedly: 80, 40, 20, 10, 5, 2.5, which is 5 halvings, so \( t = 5 \cdot 6 = 30 \) days.

17 Three ways to compound ★★★

(a) \( 8000(1.035)^{10} \approx 11284.79 \) dollars.

(b) \( 8000\left(1 + \dfrac{0.035}{4}\right)^{40} \approx 11335.27 \) dollars.

(c) \( 8000\,e^{0.35} \approx 11352.54 \) dollars.

Difference: \( 11352.54 - 11284.79 = 67.75 \) dollars.

18 Finding a transformed function ★★★

The asymptote of \( a \cdot 2^{x} + k \) is \( y = k \), so \( k = -3 \).

\( g(0) = a \cdot 1 - 3 = 5 \), so \( a = 8 \) and \( g(x) = 8 \cdot 2^{x} - 3 \).

\( g(2) = 8 \cdot 4 - 3 = 29 \) and \( g(-1) = 8 \cdot \tfrac{1}{2} - 3 = 1 \).

19 Approaching e ★★★

\( n = 12 \): \( (1.08333\ldots)^{12} \approx 2.6130 \), gap \( 2.7183 - 2.6130 \approx 0.1053 \).

\( n = 365 \): \( (1.00274)^{365} \approx 2.7146 \), gap \( \approx 0.0037 \).

The gap shrinks as \( n \) grows: the values approach \( e \). Even with interest added every day, 1 dollar grows to only about 2.71 dollars in one year, never beyond \( e \approx 2.72 \) dollars.

20 Doubling an investment ★★★

\( 5000(1.06)^{11} \approx 9491.49 < 10000 \) and \( 5000(1.06)^{12} \approx 10060.98 > 10000 \). So it takes 12 years.

Exactly: \( 1.06^{t} = 2 \), so \( t = \dfrac{\ln 2}{\ln 1.06} \approx \dfrac{0.6931}{0.05827} \approx 11.9 \) years, consistent with the table.

21 Continuous doubling time ★★★

(a) \( A(t) = 1500\,e^{0.04t} \).

(b) Divide by 1500: \( e^{0.04t} = 2 \). Take \( \ln \): \( 0.04t = \ln 2 \approx 0.6931 \), so \( t \approx 17.33 \) years.

(c) \( 1500\,e^{0.04 \cdot 17.33} = 1500\,e^{0.6932} \approx 1500 \cdot 2.0001 \approx 3000 \) dollars. It checks out.

22 Modeling app users ★★★

(a) The ratios \( 60 \div 50 = 1.2 \), \( 72 \div 60 = 1.2 \) and \( 86.4 \div 72 = 1.2 \) are equal, so the growth is exponential: \( N(t) = 50(1.2)^{t} \).

(b) \( N(6) = 50 \cdot 1.2^{6} = 50 \cdot 2.985984 \approx 149.3 \) thousand, about 149,300 users.

(c) \( N(3) = 86.4 < 100 \) and \( N(4) = 50 \cdot 2.0736 = 103.68 > 100 \): in year 4.