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Section 7.5 Exponential Models

Subsection 1. Solve power and exponential equations

Compare the procedures for solving power equations and exponential equations.

Subsubsection Examples

Example 7.54.

Solve \(~3x^{1.05} = 18~\text{.}\) Round your answer to hundredths.

Solution

This is a power equation. We divide both sides by 3 to isolate the variable, then raise both sides to the reciprocal of the exponent.

\begin{align*} (x^{1.05})^{1/1.05} \amp = 6^{1/1.05}\\ x \amp = 5.51 \end{align*}
Example 7.55.

Solve \(~3(1.05)^x = 18~\text{.}\) Round your answer to hundredths.

Solution

This is an exponential equation. We divide both sides by 3, then take logarithms.

\begin{align*} \log(1.05^x) \amp = \log 6 \amp\amp \blert{\text{Apply the third log property.}}\\ x \log 1.05\amp = \log 6\\ x \amp =\dfrac{\log 6}{\log 1.05} = 36.72 \end{align*}
Example 7.56.

Solve \(~9x^{3/5} = 36~\text{.}\) Round your answer to hundredths.

Solution

This is a power equation. We divide both sides by 9 to isolate the variable, then raise both sides to the reciprocal of the exponent.

\begin{align*} (x^{3/5})^{5/3} \amp = 4^{5/3}\\ x \amp = 10.08 \end{align*}
Example 7.57.

Solve \(~1.5(3^{x/5}) = 12~\text{.}\) Round your answer to hundredths.

Solution

This is an exponential equation. We divide both sides by 1.5, then take logarithms.

\begin{align*} \log 3^{x/5} \amp = \log 8 \amp\amp \blert{\text{Apply the third log property.}}\\ \dfrac{x}{5} \log 3 \amp = \log 8\\ x \amp =\dfrac{5\log 8}{\log 3} = 9.46 \end{align*}

Subsubsection Exercises

Solve \(~6x^{3/4}-8 = 76~\text{.}\) Round your answer to hundredths.

Answer
\(33.74\)

Solve \(~6\left(\dfrac{3}{4}\right)^x-8 = 76~\text{.}\) Round your answer to hundredths.

Answer
\(-9.17\)

Solve \(~13.2(1.36)^x = 284.8~\text{.}\) Round your answer to hundredths.

Answer
\(9.99\)

Solve \(~13.2x^{1.26} = 284.8~\text{.}\) Round your answer to hundredths.

Answer
\(11.45\)

Subsection 2. Calculate gowth and decay rates

Doubling Time and Half-Life.

If \(D\) is the doubling time for an exponential function \(P(t)\text{,}\) then

\begin{equation*} P(t) = P_0 2^{t/D} \end{equation*}

If \(H\) is the half-life for an exponential function \(Q(t)\text{,}\) then

\begin{equation*} Q(t)=Q_0 (0.5)^{t/H} \end{equation*}

Subsubsection Examples

Example 7.62.

The half-life of a cold medication in the body is 6 hours. Find its decay rate.

Solution

The decay law for the medication is

\begin{equation*} N=N_0(0.5)^{t/8} \end{equation*}

We can rewrite this expression as

\begin{equation*} N=N_0(0.5^{1/8})^t \end{equation*}

so \(b=0.5^{1/8}=0.9170\text{,}\) and \(r=1-b=0.083\text{.}\) The decay rate is 8.3%.

Example 7.63.

The growth rate of a population of badgers is 3.8% per year. Find its doubling time.

Solution

The growth law for the population is \(P=P_0(1.038)^t\text{.}\) We set \(P=2P_0\) and solve for \(t\text{.}\)

\begin{align*} 2P_0 \amp = P_0(1.038)^t \amp\amp \blert{\text{Divide both sides by}~P_0.}\\ 2 \amp = (1.038)^t \amp\amp \blert{\text{Take the log of both sides.}}\\ \log 2 \amp = t \log 1.038 \amp\amp \blert{\text{Apply the third log property.}}\\ t \amp = \dfrac{\log 2}{\log 1.038} = 18.59 \end{align*}

The doubling time is 18.59 years.

Subsubsection Exercises

The doubling time for a population is 18 years. Find its annual growth rate.

Answer

3.9%

A radioactive isotope decays by 0.04% per second. What is its half-life?

Answer

4.81 hrs

Subsection 3. Analyze graphs of exponential functions

From a graph, we can read the initial value of an exponential function and then its doubling time or hlaf-life. From there we can calculate the growth or decay law.

Subsubsection Examples

Example 7.66.

The graph shows the population, \(P\text{,}\) of a herd of llamas \(t\) years after 2000.

  1. How many llamas were there in 2000?
  2. What is the doubling time for the population?
  3. What is the annual growth rate for the population?
exponential growth
Solution
  1. The initial value of the population is given by the \(P\)-intercept of the graph, \((0,15)\text{.}\) There were 15 llamas in 2000.
  2. Look for the time when the initial llama population doubles. When \(t=4,~P=30\text{,}\) and when \(t=8,~P=60\text{,}\) so the llama population doubles every 4 years.
  3. The growth factor for the population is \(2^{1/4} = 1.189\text{,}\) so the annual growth rate is 18.9%.
Example 7.67.

Write a decay law for the graph shown below, where \(t\) is in hours and \(N\) is in milligrams.

exponential decay
Solution

The initial value is given by the vertical intercept of the graph, \((0,600)\text{,}\) so \(N_0 = 600\text{.}\)

When \(t=5,~N=300\text{,}\) so the half-life of the substance is 5 hours. Thus the decay law is \(N(t)=600(0.5)^{t/5}\text{,}\) or \(N(t)=600(0.87)^t\text{.}\)

Subsubsection Exercises

  1. Write a growth law for the population whose graph is shown, where \(t\) is in years.
  2. What is the annual growth rate for the population?
exponential growth
Answer
  1. \(\displaystyle P(t) = 10(2^{t/2.5})\)
  2. 32.0%
  1. Write a decay law for the population whose graph is shown, where \(t\) is in days.
  2. What is the daily decay rate for the population?
exponential decay
Answer
  1. \(\displaystyle P(t) = 120(0.5^{t/3})\)
  2. 20.6%