# Topic 18 Applications of Exponential and Logarithmic Functions

## 18.1 Newton’s Law of Cooling

Suppose an object with an initial temperature \(T(0)\) is placed in an environment with surrounding temperature \(T_{\text{env}}\). By Newton’s Law of Cooling, after \(t\) minutes, the temperature of the object \(T(t)\) is given by the exponential function \[ T(t)=T_{\text{env}}+(T(0)-T_{\text{env}})\,e^{-rt}, \] where \(r\) is a positive constant characteristic of the system.

A cup of coffee is brewed with a temperature 195°F and placed in a room with the temperature 60°F. The cooling constant for a cup of coffee is $ r = .09 ^{-1}$.

- After 30 minutes, what is the temperature of the coffee?
- How long it takes for the coffee to cool down to the room temperature?

## 18.2 Exponential and Logarithmic Equations

To solve an exponential or logarithmic equation, the first step is to rewrite the equation with a single exponentiation or logarithm. Then we can use the equivalent relation between exponentiation and logarithm to rewrite the equation and solve the resulting equation.

**Example 18.1 **Solve the equation \(10^{2x-1}-5=0\).

*Solution*.

- Rewrite the equation in the form \(b^u=c\): \[10^{2x-1}=5.\]
- Take logarithm of both sides and simplify: \[2x-1=\log 5.\]
- Solve the resulting equation: \[x=\frac{1}{2}(\log5+1).\]

**Example 18.2 **Solve the equation \(\log_2 x + \log_2 (x - 2) = 3\).

*Solution*.

- Rewrite the equation in the form \(\log_bu=c\): \[\log_2(x(x - 2)) = 3\]
- Rewrite the equation in the exponential form (moving the base): \[x(x-2)=2^3\]
- Solve the resulting equation \(x^2-2x-8=0\). The solutions are \(x=-2\) and \(x=4\)
- Check proposed solutions. Both \(x\) and \(x-2\) has to be positive. So \(x=-2\) is not a solution of the original equation. When \(x=4\), we have \(\log_24+\log_22=2+1=3\). So \(x=4\) is a solution.

## 18.3 Solving Compound Interest Model

**Example 18.3 **A check of $5000 was deposited in a savings account with an annual interest rate \(6\%\) which is compounded monthly. How many years will it take for the money to raise by 20%?

*Solution*.

The question tells us the following information: \(P=5000\), \(r=0.06\), \(n=12\), and \(A=5000\cdot (1+0.2)=6000\). What we want to know is the number of years \(t\). The compound interest model tells us that \(t\) satisfies the following equation: \[ 6000=5000\left(1+\frac{0.06}{12}\right)^{12t}. \] This is an exponential equation and can be solve using logarithms. \[ \begin{aligned} 5000\left(1+\frac{0.06}{12}\right)^{12t}&=6000\\ \left(1+\frac{0.06}{12}\right)^{12t}&=1.2\\ 12t\cdot \log\left(1+0.06\div 12\right)&=1.2\\ 12t&=\log(1.2)\div \log(1+0.06\div 12)\\ t&=\log(1.2)\div\log(1+0.06\div 12)\div 12\approx 3. \end{aligned} \] So it takes about 3 years for the savings to raise by 20%.

When solving exponential and logarithmic equations, you may also use the one-to-one property if both sides are powers with the same base or logarithms with the same base.

## 18.4 Practice

**Problem 18.1 **Solve the exponential equation.

- \(2^{x-1}=4\)
- \(7e^{2x}-5=58\)

**Problem 18.2 **Solve the exponential equation.

- \(3^{x^2-2x}=e^{-\ln3}\)
- \(2^{(x+1)}=3^{(1-x)}\)

**Problem 18.3 **Solve the logarithmic equation.

- \(\log_5x+\log_5(4x-1)=1\)
- \(\ln \sqrt{x+1}=1\)

**Problem 18.4 **Solve the logarithmic equation.

- \(\log_2(x+2)-\log_2(x-5)=3\)
- \(\log_3(x-5)=2-\log_3(x+3)\)

**Problem 18.5 **For the given function, find values of \(x\) satisfying the given equation.

- \(f(x)=\log_4x-2\log_4(x+1)\),\(f(x)=-1\)
- \(g(x)=\log(2-5x)+\log(-x)\),\(g(x)=1\)

**Problem 18.6 **Find intersections of the given pairs of curves.

- \(f(x)=e^{2x}\) and~ \(g(x)=e^x+12\).
- \(f(x)=\log_7\left(\frac12(x+2)\right)\) and~ \(g(x)=1-\log_7(x-3)\)

**Problem 18.7 **Using the formula \(A=P(1+\frac rn)^{nt}\) to determine how many years, to the nearest hundredth, it will take to double an investment $20,000 at the interest rate 5% compounded monthly.

**Problem 18.8 **Newton’s Law of Cooling states that the temperature \(T\) of an object at any time \(t\) satisfying the equation \(T=T_s+(T_0-T_s)e^{-rt}\),
where \(T_s\) is the the temperature of the surrounding environment,
\(T_0\) is the initial temperature of the object,
and \(r\) is positive constant characteristic of the system,
which is in units of \({\displaystyle \text{time}^{-1}}\).
In a room with a temperature of \(22 ^\circ C\), a cup of tea of \(97 ^\circ C\) was freshly brewed.
Suppose that \(r=\ln 5/20~~\text{minute}^{-1}\). In how many minutes, the temperature of the tea will be \(37 ^\circ C\)?

**Problem 18.9 **A culture of bacteria began with 1000 bacteria and grows exponentially. An hour later there were 1320 bacteria. How many hours after starting will there be 2790 bacteria?

**Problem 18.10 **A type of virus started spreading. After 10 days, 580 infected cases was confirmed. The natural log of the number of confirmed cases approximately is a linear function of the number of days after the virus started spreading. The slope of this linear function is 0.28. Suppose the pattern continues. Estimate the number of infected cases after 20 days.