Exponential Models And Applications
Apply exponential models for Queensland Year 11 Mathematical Methods (QCAA). Many real quantities grow or shrink by the same factor over equal time steps, and an exponential model captures this growth and decay.
You will learn to find the starting value and growth or decay factor, evaluate the model, solve for time using logarithms, and interpret logarithmic scales such as decibels and Richter magnitude — linking maths to science.
Every question with a fully worked solution.
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Theory
In Year 11 Mathematical Methods (QCAA, Unit 2), an exponential model \(y=a\,r^{t}\) describes quantities that grow or decay by a constant factor each period — populations, radioactive mass, and depreciating value. This page builds a model from a growth or decay rate, evaluates it, and uses logarithms to solve for the time \(t\). It also covers the decibel and Richter scales, which are logarithmic.
An exponential model has the form \(y=a\,r^{t}\), where \(a\) is the initial value (at \(t=0\)) and \(r\) is the constant growth or decay factor per period. If \(r>1\) the quantity grows; if \(0
A percentage change sets the factor: a \(p\%\) increase gives \(r=1+\dfrac{p}{100}\); a \(p\%\) decrease gives \(r=1-\dfrac{p}{100}\). To find a time — a doubling time, a half-life, or when a threshold is reached — isolate the power and take a logarithm.
Some quantities are measured on a logarithmic scale. Sound level in decibels is \(L=10\log_{10}\!\left(\dfrac{I}{I_0}\right)\), and earthquake magnitude on the Richter scale is \(R=\log_{10}\!\left(\dfrac{I}{I_0}\right)\); each step of \(1\) means a ten-fold change in intensity.
Exponential growth / decay model:
Factor from a percentage rate:
Decibel and Richter scales:
How to model and solve growth or decay
- Set up: write \(y=a\,r^{t}\) with \(a\) the initial value and \(r=1\pm\dfrac{p}{100}\) from the rate.
- Evaluate: to find a quantity, substitute the value of \(t\) and compute \(a\,r^{t}\).
- Solve for time: to find \(t\), divide to get \(r^{t}\) alone, then take \(\log_{10}\) of both sides and divide by \(\log_{10}r\).
Identify the factor — an \(8\%\) increase:
| \(r\) | \(=\) | \(1+\dfrac{8}{100}=1.08\) |
Substitute \(t=10\) and evaluate:
| \(P\) | \(=\) | \(500(1.08)^{10}\) |
| \(\approx\) | \(500\times 2.158925\) | |
| \(\approx\) | \(1079.46\) |
About \(1079\) people.
Set \(M=20\) and isolate the power:
| \(80(0.88)^{t}\) | \(=\) | \(20\) |
| \((0.88)^{t}\) | \(=\) | \(\dfrac{20}{80}=0.25\) |
Take \(\log_{10}\) of both sides and bring the index down:
| \(t\log_{10}0.88\) | \(=\) | \(\log_{10}0.25\) |
Divide by \(\log_{10}0.88\) (a negative number):
| \(t\) | \(=\) | \(\dfrac{\log_{10}0.25}{\log_{10}0.88}\) |
| \(t\) | \(\approx\) | \(\dfrac{-0.60206}{-0.05552}\) |
| \(t\) | \(\approx\) | \(10.84\) |
About \(10.8\) years.
Substitute the intensities and simplify the ratio:
| \(L\) | \(=\) | \(10\log_{10}\!\left(\dfrac{10^{-6}}{10^{-12}}\right)\) |
| \(=\) | \(10\log_{10}\!\left(10^{6}\right)\) |
Use \(\log_{10}(10^{6})=6\):
| \(L\) | \(=\) | \(10\times 6\) |
| \(=\) | \(60\) |
The sound level is \(60\) dB.
Write each intensity in index form from the definition:
| \(\dfrac{I_6}{I_0}\) | \(=\) | \(10^{6}\) |
| \(\dfrac{I_4}{I_0}\) | \(=\) | \(10^{4}\) |
Divide to compare (the \(I_0\) cancels), then subtract indices:
| \(\dfrac{I_6}{I_4}\) | \(=\) | \(\dfrac{10^{6}}{10^{4}}\) |
| \(=\) | \(10^{6-4}=10^{2}\) | |
| \(=\) | \(100\) |
The magnitude \(6\) quake is \(100\) times more intense.
Common pitfalls
Frequently asked questions
What do a and r mean in the model y = a r^t?
\(a\) is the initial value (the amount when \(t=0\)) and \(r\) is the factor you multiply by each period; \(r>1\) is growth and \(0
How do you turn a percentage rate into the factor r?
For growth of \(p\%\), \(r=1+\dfrac{p}{100}\); for decay of \(p\%\), \(r=1-\dfrac{p}{100}\). For example \(8\%\) growth gives \(r=1.08\).
How do you find the time in an exponential model?
Isolate \(r^{t}\), take \(\log_{10}\) of both sides so the exponent comes down, then divide by \(\log_{10}r\).
How does the decibel formula work?
\(L=10\log_{10}\!\left(\dfrac{I}{I_0}\right)\); substitute the intensity ratio and evaluate the base-10 logarithm.
On the Richter scale, how much stronger is magnitude 6 than magnitude 4?
Each unit is a ten-fold increase, so the difference of \(2\) means \(10^{2}=100\) times more intense.
Do exponential models here use e or natural logarithms?
No. Year 11 Methods uses \(y=a\,r^{t}\) with base-10 logarithms to solve for time; \(e\) and natural logs are Year 12.