Skip to main content
Logo image

Intermediate Algebra: Functions and Graphs

Katherine Yoshiwara

Section 7.4 Properties of Logarithms

So far we’ve seen that taking a logarithm is the opposite or inverse of raising to a power, so that we can use logarithms to solve exponential equations. And with the help of a calculator we can use base 10 or "common" logarithms to find approximate solutions to equations in base 10. Now we’ll see how to use properties of logarithms to solve exponential equations with any base.

Subsection 7.4.1 Logarithms and Exponents

Because logarithms are actually exponents, they have several properties that can be derived from the laws of exponents. Here are the laws we will need at present. \(~\alert{\text{[TK]}}\)
  1. To multiply two powers with the same base, add the exponents and leave the base unchanged.
    \begin{equation*} a^m \cdot a^n = a^{m+n} \end{equation*}
  2. To divide two powers with the same base, subtract the exponents and leave the base unchanged.
    \begin{equation*} \frac{a^m}{a^n} = a^{m-n} \end{equation*}
  3. To raise a power to a power, keep the same base and multiply the exponents.
    \begin{equation*} \left(a^m\right)^n = a^{mn} \end{equation*}
Each of these laws corresponds to a property of logarithms.

Properties of Logarithms.

If \(x,~y,\) and \(b\gt 0,\) and \(b\ne 1\text{,}\) then
  1. \(\displaystyle \log_b {(xy)} = \log_b {(x)} + \log_b {(y)}\)
  2. \(\displaystyle \log_b {\left(\dfrac{x}{y}\right)} = \log_b {(x)} - \log_b {(y)}\)
  3. \(\displaystyle \log_b {(x^k)} = k\log_b {(x)} \)
To help you see why these properties work, study the examples below, keeping in mind that a logarithm is an exponent.
  1. Property (1):
    \begin{alignat*}{4} \log_{2}{(32)} \amp=\log_{2}{(4 \cdot 8)} \amp\amp={}\amp\log_2 (4) \amp+\log_2 (8) \amp \quad\text{ because } 2^{\alert{5}} \amp= 2^{\alert{2}} \cdot 2^{\alert{3}}\\ \alert{5} \amp\amp\amp = \amp\alert{2} \amp+ \alert{3} \amp 32 \amp = 4 \cdot 8 \end{alignat*}
  2. Property (2):
    \begin{alignat*}{4} \log_{2}{(8)} \amp=\log_{2}{\left(\frac{16}{2}\right)} \amp\amp={}\amp\log_2{(16)}\amp-\log_2 {(2)} \amp \quad\text{ because } 2^{\alert{3}} \amp= \frac{2^{\alert{4}}}{2^{\alert{1}}} \\ \alert{3} \amp\amp\amp = \amp\alert{4} \amp- \alert{1} \amp 8 \amp = \frac{16}{2} \end{alignat*}
  3. Property (3):
    \begin{alignat*}{3} \log_{2}{(64)} \amp=\log_{2}{(4^3)} \amp\amp=3\log_2 (4) \amp \quad\text{ because }\left(2^{\alert{2}}\right)^{\alert{3}} \amp= 2^{\alert{6}}\\ \alert{6} \amp\amp\amp = \alert{3}\cdot \alert{2} \amp (4)^{\alert{3}} \amp = 64 \end{alignat*}

Checkpoint 7.4.1. QuickCheck 1.

Which of the statements below are true?
  1. \(\displaystyle \log {(8+y)}=\log {(8)}+\log {(y)}\)
  2. \(\displaystyle \log {(8y)}=\log {(8)}+\log {(y)}\)
  3. \(\displaystyle \log {(8y)}=\log {(8)} \cdot \log {(y)}\)
  4. \(\displaystyle \log \left(\dfrac{8}{y}\right)~=\dfrac{\log {(8)}}{\log {(y)}}\)

Subsection 7.4.2 Using the Properties of Logarithms

Of course, these properties are useful not so much for computing logs but rather for simplifying expressions that contain variables. We will use them to solve exponential equations. But first, we will practice applying the properties. \(~\alert{\text{[TK]}}\)

Example 7.4.2.

Write \(~~\log_{b}{(\sqrt{xy})}~~\) as an expression in simpler logs.

Solution.

First, we write \(\sqrt{xy}~~\) using a fractional exponent:
\begin{align*} \log_{b}{(xy)} \amp = \log_{b}{((xy)^{1/2})} \amp \amp \blert{\text{Apply Property 3.}}\\ \amp = \frac{1}{2}\log_{b}{(xy)} \amp \amp \blert{\text{Apply Property 1 to the product}~xy.}\\ \amp = \frac{1}{2}(\log_{b}{(x)} + \log_{b}{(y)}) \end{align*}
Thus, \(~~\log_{b}{(\sqrt{xy})} = \dfrac{1}{2}(\log_{b}{(x)} + \log_{b}{(y)})\text{.}\)

Checkpoint 7.4.3. Practice 1.

Write \(~\log_{b}{\left(\dfrac{x}{y^2}\right)}~\) as an expression in simpler logs.
Answer.
\(\log_b {(x}) - 2\log_b {(y)}\)
We can also use the properties of logarithms to combine sums and differences of logarithms into one logarithm.

Example 7.4.4.

Express \(~~3(\log_b {(x)} - \log_b {(y)})\) as a single logarithm with a coefficient of \(1\text{.}\)

Solution.

We begin by applying Property (2) to combine the logs.
\begin{align*} 3(\log_b {(x)} - \log_b {(y)}) \amp = 3 \log_{b}{\left(\dfrac{x}{y}\right)} \amp \amp \blert{\text{Apply Property 3.}}\\ \amp =\log_{b}{\left(\dfrac{x}{y}\right)^3} \end{align*}
Thus, \(3(\log_b {(x)} - \log_b {(y)}) = \log_{b}{\left(\dfrac{x}{y}\right)^3}\text{.}\)

Caution 7.4.5.

Be careful when using the properties of logarithms! Compare the statements below:
  1. \(\log_{b}{(2x)} = \log_{b}{(2)} + \log_{b}{(x)}~~~~ \text{ by Property 1,}\)
    but
    \(\log_{b}{(2 + x)} \ne \log_{b}{(2)} + \log_{b}{(x)}\)
  2. \(\log_{b}{\left(\dfrac{x}{5}\right)}= \log_b {(x)} - \log_b {(5)}~~~~ \text{ by Property 2,}\)
    but
    \(\log_{b}{\left(\dfrac{x}{5}\right)} \ne \dfrac{\log_b {(x)}}{\log_b {(5)}}\)

Checkpoint 7.4.6. QuickCheck 2.

Decide whether each statement is true or false.
  1. The log of a product is equal to the sum of the logs of the factors.
  2. The log of a quotient is the equal to the difference of the logs.
  3. The log of a power is equal to the exponent times the log of the base.
  4. The distributive law applies to the operation of taking a logarithm of a sum or difference.

Subsection 7.4.3 Solving Exponential Equations

By using Property (3), we can now solve exponential equations in which the base is not 10. For example, to solve the equation
\begin{equation*} 5^x = 7 \end{equation*}
we could rewrite the equation in logarithmic form to obtain the exact solution
\begin{equation*} x = \log_{5}{(7)} \end{equation*}
However, we cannot evaluate \(\log_{5}{(7)}\text{;}\) there is no "log base 5 button" on the calculator. If we want a decimal approximation for the solution, we begin by taking the base 10 logarithm of both sides, even though the base of the power is not 10. This gives us
\begin{equation*} \log_{10}{(5^x)} = \log_{10}{(7)} \end{equation*}
Then we use Property (3) to rewrite the left side as
\begin{equation*} x \log_{10}{(5)} = \log_{10}{(7)} \end{equation*}
and divide both sides by \(\log_{10}{(5)}\) to get
\begin{equation*} x = \frac{\log_{10}{(7)}}{\log_{10}{(5)}} \end{equation*}
On your calculator, enter the sequence
LOG \(7\) ) ÷ LOG \(5\) ) ENTER
to find that \(x \approx 1.2091\text{.}\)

Note 7.4.7.

Note how using Property (3) allows us to solve the equation above: The variable, \(x\text{,}\) is no longer in the exponent, and it is multiplied by a constant, \(\log_{10}{(5)}\text{.}\)

Caution 7.4.8.

Do not confuse the expression \(~\dfrac{\log_{10}{(7)}}{\log_{10}{(5)}}~\) with \(~\log_{10}{\left(\dfrac{7}{5}\right)}~\text{;}\) they are not the same! Although
\begin{equation*} \log_{10}{\left(\dfrac{7}{5}\right)}= \log_{10}{(1.4)} \approx 0.1416 \end{equation*}
We cannot simplify \(~\dfrac{\log_{10}{(7)}}{\log_{10}{(5)}}~\text{;}\) we must evaluate it as
\begin{equation*} \log_{10}{(7)} \div \log_{10}{(5)} \approx 1.2091 \end{equation*}

Example 7.4.9.

Solve \(~~1640 = 80 \cdot 6^{0.03x}\) \(~\alert{\text{[TK]}}\)

Solution.

First we divide both sides by 80 to isolate the power.
\begin{equation*} 20.5 = 6^{0.03x} \end{equation*}
Next, we take the base 10 logarithm of both sides of the equation and use Property (3) to get
\begin{align*} \log_{10}{(20.5)} \amp = \log_{10}{(6^{0.03x})} \amp \amp \blert{\text{Apply Property 3.}}\\ \amp = 0.03x \log_{10}{(6)} \end{align*}
On the right side of the equation, \(x\) is multiplied by two constants, \(0.03\) and \(\log_{10}{(6)}\text{.}\) So, to solve for \(x\) we must divide both sides of the equation by \(0.03 \log_{10}{(6)}\text{.}\) We use a calculator to evaluate the answer:
\begin{equation*} x = \frac{\log_{10}{(20.5)}}{0.03 \log_{10}{(6)}}\approx 56.19 \end{equation*}
(On your calculator, remember to enclose the entire denominator, \(0.03 \log_{10}{(6)}\text{,}\) in parentheses.)

Caution 7.4.10.

In the previous example, do not try to simplify
\begin{equation*} 80 \cdot 6^{0.03x} \rightarrow 480^{0.03x}~~ \blert{\text{ Incorrect!}} \end{equation*}
Remember that the order of operations tells us to compute the power \(6^{0.03x}\) before multiplying by 80. To solve the equation, we must first isolate the power with the variable exponent.

Checkpoint 7.4.11. Practice 3.

Solve \(5(1.2)^{2.5x} = 77\)
Answer.
\(x=\dfrac{\log {(15.4)}}{2.5\log {(1.2)}}\approx 5.999 \)
By using the properties of logarithms, we can now solve equations that arise in exponential growth and decay models, no matter what base the exponential function uses.

Example 7.4.12.

The population of Silicon City was 6500 in 1990 and has been tripling every 12 years. When will the population reach 75,000?

Solution.

The population of Silicon City grows according to the formula
\begin{equation*} P(t) = 6500 \cdot 3^{t/12} \end{equation*}
where \(t\) is the number of years after. We want to find the value of \(t\) for which \(P(t) = 75,000\text{;}\) that is, we want to solve the equation
\begin{equation*} 6500 \cdot 3^{t/12} = 75,000 \end{equation*}
To isolate the power, we divide both sides by 6500 to get
\begin{equation*} 3^{t/12} = \dfrac{150}{13} \end{equation*}
Now we can take the base 10 logarithm of both sides and solve for \(t\text{.}\)
\begin{align*} \log_{10}{(3^{t/12})} \amp = \log_{10}{\left(\frac{150}{13}\right)} \amp\amp \blert{\text{Apply Property (3).}}\\ \frac{t}{12}\log_{10}{(3)} \amp= \log_{10}{\left(\frac{150}{13}\right)} \amp\amp \blert{\text{Divide by }\log_{10}{(3)}\text{; multiply by 12.}}\\ t \amp = \frac{12 \left(\log_{10}{(\frac{150}{13}}\right)}{\log_{10}{(3)}} \end{align*}
We use a calculator to evaluate the answer, \(t \approx 26.71\text{.}\) The population of Silicon City will reach 75,000 about 27 years after 1990, or in 2017.

Checkpoint 7.4.13. Practice 4.

The volume of traffic on U.S. highways is growing by 2.7% per year. (Source: Time, Jan. 25, 1999)
  1. Write a formula for the volume, \(V\text{,}\) of traffic as a function of time, using \(V_0\) for the current volume.
  2. How long will it take the volume of traffic to double?
    Hint: Find the value of \(t\) that gives \(V = 2V_0\text{.}\)
Answer.
  1. \(\displaystyle V(t) = V_0(1.027)^t\)
  2. about 26 years

Checkpoint 7.4.14. QuickCheck 3.

The first step for solving each equation below is incorrect. Replace it with the correct step.
  1. \begin{align*} 6(5^{3t})\amp=20\\ \color{gray}{(\log 6)(\log 5^{3t})}\amp=\color{gray}{\log {20}}\amp\amp \blert{\text{(Wrong!)}} \end{align*}
  2. \begin{align*} t \log {(8)}\amp=\log{(53)}\\ \color{gray}{t}\amp=\color{gray}{\log {\dfrac{53}{8}}}\amp\amp \blert{\text{(Wrong!)}} \end{align*}
  3. \begin{align*} 12(10)^{-0.06t}\amp=192\\ \color{gray}{120^{-0.06t}}\amp=\color{gray}{192}\amp\amp \blert{\text{(Wrong!)}} \end{align*}
  4. \begin{align*} \log {(x)} +\log {(x-2)}\amp=3 \\ \color{gray}{x + (x-2)}\amp=\color{gray}{10^3}\amp\amp \blert{\text{(Wrong!)}} \end{align*}

Subsection 7.4.4 Solving Formulas

We can also solve formulas involving exponential or logarithmic expressions for one variable in terms of the others.

Example 7.4.15.

Solve \(~~P = Cb^{kt}~~\) for \(t\text{.}\) (\(C\) and \(k \ne 0\text{.}\))

Solution.

First, we divide both sides by \(C\) to isolate the power.
\begin{equation*} b^{kt} = \dfrac{P}{C} \end{equation*}
Then we write the exponential equation in logarithmic form:
\begin{align*} kt \amp = \log_b {\left(\dfrac{P}{C}\right)} \amp\amp \blert{\text{Divide both sides by}~k.}\\ t \amp = \dfrac{1}{k}\log_b {\left(\dfrac{P}{C}\right)} \end{align*}

Checkpoint 7.4.16. Practice 5.

Solve \(~N=N_0 \log_b{(ks)}\) for \(s\text{.}\)
Hint.
Follow the steps:
  1. Divide both sides by \(N_0\text{.}\)
  2. Rewrite in exponential form.
  3. Solve for \(s\text{.}\)
Answer.
\(s=\dfrac{1}{k}b^{N/N_0}\)
One final note about finding approximate values for logarithms. As we saw earlier, the solution of the equation \(~5^x = 7~\) is \(~x = \log_5 {(7)}\text{,}\) but by taking the log base 10 of both sides of the equation we obtained the solution in the form
\begin{equation*} x = \frac{\log_{10}{(7)}}{\log_{10}{(5)}} \end{equation*}
Because these two expressions for the solution must be equal, we have
\begin{equation*} \log_5 {(7)} = \frac{\log_{10}{(7)}}{\log_{10}{(5)}} \end{equation*}
This equation is an example of the change of base formula.

Change of Base Formula.

For positive values of \(a, b\text{,}\) and \(x\text{,}\) and \(~a, b \not= 1\text{,}\)
\begin{equation*} \blert{\log_a {(x)} = \frac{\log_{b}{(x)}}{\log_{b}{(a)}}} \end{equation*}
We can use the change of base formula to convert any logarithm to base 10, or any other convenient base.

Exercises 7.4.5 Problem Set 7.4

Warm Up

1.
Convert each equation to logarithmic form.
  1. \(\displaystyle 8^{-1/3} = \dfrac{1}{2}\)
  2. \(\displaystyle 5^x = 46\)
2.
Convert each equation to exponential form.
  1. \(\displaystyle \log_{10} {(C)} = -4.5\)
  2. \(\displaystyle \log_m {(n)} = 5\)
Exercise Group.
For Problems 3-8, follow the instructions, then state which property of logarithms each problem illustrates.
3.
  1. Simplify \(10^2 \cdot 10^6\)
  2. Compute \(\log {(10^2)},~\log {(10^6)},~\) and \(~\log {(10^2 \cdot 10^6)}\)
4.
  1. Simplify \(\dfrac{10^9}{10^6}\)
  2. Compute \(\log {(10^9)},~\log {(10^6)},~\) and \(~\log {\left(\dfrac{10^9}{10^6}\right)}\)
5.
  1. Simplify \(b^8 \cdot b^5\)
  2. Compute \(\log_b {(b^8)},~\log_b {(b^5)},~\) and \(~\log_b {\left(\dfrac{b^8}{b^5}\right)}\)
6.
  1. Simplify \(b^4 \cdot b^3\)
  2. Compute \(\log_b {(b^4)},~\log_b {(b^3)},~\) and \(~\log_b {(b^4 \cdot b^3)}\)
7.
  1. Simplify \((10^3)^5\)
  2. Compute \(\log {((10^3)^5)},~\) and \(~\log {(10^3)},~\)
8.
  1. Simplify \((b^2)^6 \)
  2. Compute \(\log_b {((b^2)^6)},~\) and \(~\log_b {(b^2)}\)
Exercise Group.
For Problems 9-12, evaluate each expression. Which (if any) are equal?
9.
  1. \(\displaystyle \log_2{(4\cdot 8)} \)
  2. \(\displaystyle \left(\log_2 {(4)}\right)\left(\log_2 {(8)}\right) \)
  3. \(\displaystyle \log_2 {(4)} + \log_2 {(8)} \)
10.
  1. \(\displaystyle \log_3{(27^2}) \)
  2. \(\displaystyle \left(\log_3 {(27)}\right)^2 \)
  3. \(\displaystyle \log_3 {(27)} + \log_3 {(27)} \)
11.
  1. \(\displaystyle \log_{10}{\left(\dfrac{240}{10} \right)} \)
  2. \(\displaystyle \dfrac{\log_{10} {(240)}}{\log_{10}{(10)}} \)
  3. \(\displaystyle \log_{10} {(240-10)} \)
12.
  1. \(\displaystyle \log_{10}{\left(\dfrac{1}{2}\cdot 80 \right)} \)
  2. \(\displaystyle \dfrac{1}{2} \log_{10} {(80)} \)
  3. \(\displaystyle \log_{10} {(\sqrt{80})} \)

Skills Practice

Exercise Group.
For Problems 13-16, use the properties of logarithms to write each expression in terms of simpler logarithms. All variable expressions denote positive numbers.
13.
  1. \(\displaystyle \log_b {(2x)} \)
  2. \(\displaystyle \log_b {\left(\dfrac{x}{2}\right)} \)
  3. \(\displaystyle \log_b {(x^3)} \)
  4. \(\displaystyle \log_b {\left(\sqrt[3]{x^2}\right)} \)
14.
  1. \(\displaystyle \log_b {\left(\dfrac{2x}{x-2}\right)} \)
  2. \(\displaystyle \log_b {\left(x(2x+3)\right)} \)
  3. \(\displaystyle \log_4 {(x^2y^3)} \)
  4. \(\displaystyle \log_3 {\left(\sqrt[3]{x^2+1}\right)} \)
15.
  1. \(\displaystyle \log_3 {(3x^4)} \)
  2. \(\displaystyle \log_5 {\left(1.1^{1/t}\right)} \)
  3. \(\displaystyle \log_b {\left((4b)^t\right)} \)
  4. \(\displaystyle \log_2 {\left(5(2^x)\right)} \)
16.
  1. \(\displaystyle \log_{10} {\left(\sqrt{\dfrac{2L}{R^2}}\right)} \)
  2. \(\displaystyle \log_{10} {\left(2\pi\sqrt{\frac{l}{g}}\right)} \)
Exercise Group.
For Problems 17-20, combine into one logarithm and simplify. Assume all expressions are defined.
17.
  1. \(\displaystyle \log_b {(8)} - \log_b {(2)}\)
  2. \(\displaystyle 2 \log_4 {(x)} + 3\log_4 {(y)}\)
18.
  1. \(\displaystyle \log_b {(5)} + \log_b {(2)}\)
  2. \(\displaystyle \dfrac{1}{4} \log_5 {(x)} - \dfrac{3}{4} \log_5 {(y)}\)
19.
  1. \(\displaystyle \log {(2x)} + 2\log {(x)} -\log{(\sqrt{x})} \)
  2. \(\displaystyle \log{(t^2-16)}-\log{(t+4)} \)
20.
  1. \(\displaystyle \log {(x^2)} + \log {(x^3)} - 5\log {(x)}\)
  2. \(\displaystyle \log{(x^2-x)}-\log{(\sqrt{x^3})} \)
Exercise Group.
For Problems 21-24, solve for the unknown value.
21.
\(\log_5 {(y)} = -2\)
22.
\(\log_b {(625)} = 4\)
23.
\(\log_b {(10)} = \dfrac{1}{2}\)
24.
\(\log_5 {(9-4x)} = 3\)
Exercise Group.
For Problems 25–30, use base 10 logarithms to solve. Round your answers to hundredths.
25.
\(2^x=7 \)
26.
\(4^{x^2}=15 \)
27.
\(4.26^{-x}=10.3 \)
28.
\(25\cdot 3^{2.1x}=47 \)
29.
\(3600=20\cdot 8^{-0.2x} \)
30.
\(3^{2x-3} = 4529 \)
Exercise Group.
For Problems 31 and 32, use the three logs below to find the value of each expression.
\begin{equation*} \log_b {(2)} = 0.6931, ~~\log_b {(3)} = 1.0986, ~~\log_b {(5)} = 1.6094 \end{equation*}
(Hint: For example, \(\log_b {(15)} = \log_b {(3)} + \log_b {(5)}\text{.}\))
31.
  1. \(\displaystyle \log_b {(6)}\)
  2. \(\displaystyle \log_b {\left(\dfrac{2}{5}\right)} \)
32.
  1. \(\displaystyle \log_b {(75)}\)
  2. \(\displaystyle \log_b {(\sqrt{50})} \)
Exercise Group.
In Problems 33-35, which expressions are equivalent?
33.
  1. \(\displaystyle \log {\left(\dfrac{x}{4}\right)}\)
  2. \(\displaystyle \dfrac{\log {(x)}}{\log {(4)}}\)
  3. \(\displaystyle \log{(x-4)}\)
  4. \(\displaystyle \log {(x)} - \log {(4)} \)
34.
  1. \(\displaystyle \log{(x+2)}\)
  2. \(\displaystyle \log {(x)}+\log {(2)}\)
  3. \(\displaystyle \log{(2x)}\)
  4. \(\displaystyle \left(\log {(2)}\right) \left(\log {(x)}\right)\)
35.
  1. \(\displaystyle \log {(x^3)}\)
  2. \(\displaystyle \left(\log {(3)}\right)\left(\log {(x)}\right)\)
  3. \(\displaystyle 3\log {(x)}\)
  4. \(\displaystyle \log {(3^x)} \)
36.
  1. Explain why \(~\log {\left(\dfrac{1}{x}\right)}~\) is the same as \(~-\log {(x)}\text{.}\)
  2. Explain why \(~\log {(x^2)}~\) is the same as \(~2 \log {(x)}\text{.}\)

Applications

37.
If hamburger is allowed to thaw at \(50\degree\)F, Salmonella bacteria grows at a rate of 9% per hour.
  1. Write a formula for the amount of Salmonella present after \(t\) hours, if the initial amount is \(S_0\text{.}\)
  2. Health officials advise that the amount of Salmonella initially present in hamburger should not be allowed to increase by more than 50%. How long can hamburger be safely left to thaw at \(50\degree\)F?
38.
Starting in 1998, the demand for electricity in Ireland grew at a rate of 5.8% per year. In 1998, 20,500 gigawatts were used. (Source: Electricity Supply Board of Ireland)
  1. Write a formula for electricity demand in Ireland as a function of time.
  2. If demand continues to grow at the same rate, when would it reach 30,000 gigawatts?
39.
The concentration of a certain drug injected into the bloodstream decreases by 20% each hour as the drug is eliminated from the body. The initial dose creates a concentration of 0.7 milligrams per milliliter.
  1. Write a function for the concentration of the drug as a function of time.
  2. The minimum effective concentration of the drug is 0.4 milligrams per milliliter. When should the second dose be administered?
  3. Graph the function, and label the point on the graph that illustrates your answer to part (b).
40.
A small pond is tested for pollution, and the concentration of toxic chemicals is found to be 80 parts per million. Clean water enters the pond from a stream, mixes with the polluted water, then leaves the pond so that the pollution level is reduced by 10% each month.
  1. Write a function for the concentration of toxic chemicals as a function of time.
  2. How long will it be before the concentration of toxic chemicals reaches a safe level of 25 parts per million?
  3. Graph the function, and label the point on the graph that illustrates your answer to part (b).
41.
Sodium-24 is a radioactive isotope that is used in diagnosing circulatory disease. It decays into stable isotopes of sodium at a rate of 4.73% per hour.
  1. Technicians inject a quantity of sodium-24 into a patient’s bloodstream. Write a formula for the amount of sodium-24 present in the bloodstream as a function of time.
  2. How long will it take for 75% of the isotope to decay?
42.
The population of Afghanistan is growing at 2.6% per year.
  1. Write a formula for the population of Afghanistan as a function of time.
  2. In 2005, the population of Afghanistan was 29.9 million. At the given rate of growth, how long would it take the population to reach 40 million?
Exercise Group.
For Problems 43-46, solve the formula for the specified variable.
43.
\(N = N_0 a^{kt}\text{,}\) for \(k\)
44.
\(Q = Q_0 b^{t/2}\text{,}\) for \(t\)
45.
\(A = A_0(10^{kt}-1)\text{,}\) for \(t\)
46.
\(w = pv^q\text{,}\) for \(q\)
You have attempted of activities on this page.
PrevTopNext