Math · College algebra · Worked example

Find when continuous growth reaches a target

A town of $12, 000$ people grows continuously at $2.5 %$ a year. When will it reach $20, 000$ ?

12000 e^{0.025 t} = 20000

Isolate the power

Divide both sides by $12, 000$ . The power must stand alone before a logarithm can bring $t$ down, and $20, 000 / 12, 000 = 5 / 3$ .

e^{0.025 t} = \frac{5}{3}

Take $ln$ of both sides

$ln$ undoes the base $e$ , since $ln (e^{u}) = u$ for every $u$ . The exponent comes down whole.

0.025 t = ln \frac{5}{3}

Solve for $t$

Divide by the rate. $ln (5 / 3) \approx 0.5108$ , and $0.5108 / 0.025 \approx 20.43$ .

t = \frac{ln \frac{5}{3}}{0.025}

Round and check

$t \approx 20.4 years$ . The rule of $70$ gives a doubling time of $70 \div 2.5 = 28 years$ , and growing by $2 / 3$ should take a good part of one doubling, so $20 years$ is reasonable. $12, 000 e^{0.025 \times 20.43}$ $\approx 20, 000$ .

t \approx 20.4

Result

After about $20.4 years$ .

t = \frac{ln \frac{5}{3}}{0.025} \approx 20.4

Your turn

1. One like the example

A town of $8000$ grows continuously at $3 %$ a year. When does it reach $12, 000$ ?

Show the answer and explanation

After about $13.5 years$ .

Divide by $8000$ to get $e^{0.03 t} = 1.5$ . Then $0.03 t = ln 1.5 \approx 0.4055$ , so $t \approx 13.5 years$ . Growing by half should take less than the doubling time of about $23 years$ , and it does.

\begin{array}{l} 8000 e^{0.03 t} = 12000 \\ e^{0.03 t} = 1.5 \\ 0.03 t = ln 1.5 \\ t = \frac{ln 1.5}{0.03} \\ t \approx 13.5 \end{array}

2. A variation

A culture grows continuously from $500$ cells to 2000 cells in $6 hours$ . Find its rate $k$ per hour.

Show the answer and explanation

$k = ln 4 / 6$ $\approx 0.231 per hour$ .

Here the time is known and the rate is not. $500 e^{6 k} = 2000$ gives $e^{6 k} = 4$ , so $6 k = ln 4 \approx 1.386$ and $k \approx 0.231$ . Quadrupling is two doublings in $6 hours$ , one every $3 hours$ , and $ln 2 / 0.231 \approx 3$ .

\begin{array}{l} 500 e^{6 k} = 2000 \\ e^{6 k} = 4 \\ 6 k = ln 4 \\ k = \frac{ln 4}{6} \\ k \approx 0.231 \end{array}

3. Somewhere new

A car worth $$ 30, 000$ loses value continuously at $15 %$ a year. When is it worth $$ 12, 000$ ?

Show the answer and explanation

After about $6.1 years$ .

Decay makes the rate negative, so the model is $30, 000 e^{- 0.15 t}$ . Setting it to $12, 000$ gives $e^{- 0.15 t} = 0.4$ , so $- 0.15 t$ $= ln 0.4$ $\approx - 0.9163$ and $t \approx 6.1 years$ . Both negatives cancel, so the time comes out positive.

\begin{array}{l} 30000 e^{- 0.15 t} = 12000 \\ e^{- 0.15 t} = 0.4 \\ - 0.15 t = ln 0.4 \\ t = \frac{ln 0.4}{- 0.15} \\ t \approx 6.1 \end{array}

Keep exploring

Open the graph and find where the curve $y = 12000 e^{0.025 t}$ meets the line $y = 20, 000$ . They cross at $t \approx 20.4$ .

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Isolate the power

Take lnln of both sides

Solve for tt

Round and check

Result

Your turn

1. One like the example

After about 13.5 years13.5 years.

2. A variation

k=ln4⁢/6k=ln4/6 ≈0.231 per hour≈0.231 per hour.

3. Somewhere new

After about 6.1 years6.1 years.

Keep exploring

Study it, then try it

Take $ln$ of both sides

Solve for $t$

After about $13.5 years$ .

$k = ln 4 / 6$ $\approx 0.231 per hour$ .

After about $6.1 years$ .