Some Applications Of Exponential Equations
- Population Growth and Decline
- Radio-Carbon Dating
- Compound Interest
Applications Of Exponential Equations
Most applications of exponential equations will be dealing with exponential functions of the form $$A(t)=A_0 e^{rt}.$$ Some problems will give you the initial amount $A_0$ of what is being studied (substance, people, money, etc.). and a rate of growth or decline $r.$ If it's a growth problem, $r$ is positive. If it's decay or decline, $r$ is negative.
Much of the time, we want to know the time at which them amount $A$ reaches some level. This will involve solving an exponential equation.
Modelling Population Growth
Example: The safe level of psychrotrophic bacteria in a gallon of milk is 100 units. In a refrigerator set at 38°F, the number of units of these bacteria in a gallon of skim milk is approximated by the exponential function $$B(t)=4.0e^{0.24t},$$ where $t$ is the time in days. How long will it take the bacterial colony to grow to 160 units? Round your answer to the nearest hundredth.
Radio Carbon Dating
Example: Carbon-14 decays continuously at the rate of $0.01245\%$ per year. An archaeologist named has determined that only $3\%$ of the original carbon-14 from a plant specimen remains. Estimate the age of this specimen.
Inflation
Example Suppose that in the country of ia that prices will double in 4 years at the current rate of inflation. What is the current rate of inflation in that country? Assume that the effect of inflation is continuous. Round your answer to the nearest hundredth of a percent.
Compound Interest (Discrete)
Compound interest has two flavors. The first is the usual method which banks use to compute compound interest: $$A=P\left(1+\frac{r}{n}\right)^{nt},$$ where $t$ is the time in years, $A$ is the amount of money, $P$ is the principal amount, $r$ is the interest rate, and $n$ is the number of compounding periods per year.
Compound Interest (Continuous)
The second flavor is when we allow the compounding to happen so often that it becomes continuous: $$A=Pe^{rt}.$$ Note #1: this is essentially exponential growth, which again models things like growing rabbit populations which "compound continuously."
Note #2: this is where the number $e$ "comes from." The above formula is essentially the discrete compound formula with "tons and tons" (i.e. infinite) compounding periods.
Compound Interest
Example: Suppose invests $\$5000$ at $2\%$ with interest compounded monthly. How long will it take for this investment to double its value? Round your answer to the nearest tenth.
Example: Suppose invests $\$5000$ at $2\%$ with interest compounded continuously. How long will it take for this investment to double its value? Round your answer to the nearest tenth.
The Number $e$ and Compound Interest
Example: Suppose $\$1$ is invested at $100\%$ with interest. How much will you have at the end of 1 year if it's compounded
- yearly?
- semi-annually?
- quarterly?
- monthly?
- every second?
- every nano-second?
Some Applications Of Logarithmic Equations
- pH of a chemical solution
- Logarithmic Scales (Richter Scale, Decibel Scale)
- Finance
The pH of a Chemical Solution
Example: The formula $\mbox{pH}=−\log \mbox{H}^+$ expresses the $\mbox{pH}$ of a solution in terms of its hydrogen ion concentration $\mbox{H}^+.$ A certain brand of shampoo has a $\mbox{pH}$ of 8.72. What is the $\mbox{H}^+$ concentration in moles per liter?
The Richter Scale
Example: Seismologists use the Richter scale to measure the magnitude of earthquakes. The equation $$R=\log\frac{A}{a}$$ compares the amplitude $A$ of the shock wave of an earthquake to the amplitude $a$ of a reference shock wave of minimal intensity. The amplitude of the September 20th, 1965, earthquake in Mexico City was 20000000 times the reference amplitude. Calculate the magnitude of this earthquake on the Richter scale. Round your answer to the nearest hundredth.
A Finance Formula
Example: The number $n$ of monthly payments of amount $P$ required to completely pay off a loan of amount $A$ borrowed at interest rate $R$ is given by the formula $$n=-\frac{\log\left(1-\frac{AR}{12P}\right)}{\log\left(1+\frac{R}{12}\right)}.$$ Determine the number of monthly car payments of $\$300$ required to pay off a $\$15000$ car loan when the interest rate is $10\%.$ Round your answer to the nearest number of payments.