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Section 7.4 Properties of Logarithms

Subsection Logarithms are 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.

  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 one of three properties 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. \(\displaystyle \log_b {\dfrac{x}{y}} = \log_b {x} - \log_b {y}\)
  3. \(\displaystyle \log_b {x^k} = k\log_b {x} \)

Study the examples below, keeping in mind that a logarithm is an exponent.

  1. Property (1):

    \begin{align*} \log_{2}{32} \amp=\log_{2}{(4 \cdot 8)} \amp\amp=\log_2 4 +\log_2 8 \amp \text{ because } 2^{\alert{5}} \amp= 2^{\alert{2}} \cdot 2^{\alert{3}}\\ \alert{5} \amp\amp\amp = \alert{2} + \alert{3} \amp 32 \amp = 4 \cdot 8 \end{align*}

  2. Property (2):

    \begin{align*} \log_{2}{8} \amp=\log_{2}{\frac{16}{2}} \amp\amp=\log_2{16}-\log_2 {2} \amp \text{ because } 2^{\alert{3}} \amp= \frac{2^{\alert{4}}}{2^{\alert{1}}} \\ \alert{3} \amp\amp\amp = \alert{4} - \alert{1} \amp 8 \amp = \frac{16}{2} \end{align*}

  3. Property (3):

    \begin{align*} \log_{2}{64} \amp=\log_{2}{(4)^3} \amp\amp=3\log_2 4 \amp \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{align*}

Subsection 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.

Example 7.62.

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, \(~~\log_{b}\sqrt{xy} = \dfrac{1}{2}(\log_{b}{x} + \log_{b}{y})\text{.}\)

We can also use the properties of logarithms to combine sums and differences of logarithms into one logarithm.

Example 7.64.

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 7.65.

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

    \(\log_{b}{\left(\dfrac{x}{5}\right)} \ne \dfrac{\log_b x}{\log_b 5}\)

Subsection 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 7.68.

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 7.69.

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 7.70.

Solve \(~~1640 = 80 \cdot 6^{0.03x}\)

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 denominator, \(0.03 \log_{10}{6}\text{,}\) in parentheses.)

Caution 7.71.

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.

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 7.73.

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.

Subsection Solving Formulas

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

Example 7.76.

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 {\dfrac{P}{C}} \amp\amp \blert{\text{Divide both sides by}~k.}\\ t \amp = \dfrac{1}{k}\log_b {\dfrac{P}{C}} \end{align*}

Exercises 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. \(\displaystyle \log_m n = 5\)

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 \(~\log {\dfrac{10^9}{10^6}}\)

5.

  1. Simplify \(b^8 \cdot b^5\)

  2. Compute \(\log_b {b^8},~\log_b {b^5},~\) and \(~\log_b {\dfrac{b^8}{b^5}}\)

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 \(~\log_b {b^2}\)

For Problems 9-12, evaluate each expression. Which (if any) are equal?

9.

  1. \(\displaystyle \log_2(4\cdot 8) \)

  2. \(\displaystyle (\log_2 4)(\log_2 8) \)

  3. \(\displaystyle \log_2 4 + \log_2 8 \)

10.

  1. \(\displaystyle \log_3(27^2) \)

  2. \(\displaystyle (\log_3 27)^2 \)

  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

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. \(\displaystyle \log_b\dfrac{x}{2} \)

  3. \(\displaystyle \log_b x^3 \)

  4. \(\displaystyle \log_b \sqrt[3]{x^2} \)

14.

  1. \(\displaystyle \log_b\dfrac{2x}{x-2} \)

  2. \(\displaystyle \log_b x(2x+3) \)

  3. \(\displaystyle \log_4 x^2y^3 \)

  4. \(\displaystyle \log_3 \sqrt[3]{x^2+1} \)

15.

  1. \(\displaystyle \log_3 (3x^4) \)

  2. \(\displaystyle \log_5 1.1^{1/t} \)

  3. \(\displaystyle \log_b (4b)^t \)

  4. \(\displaystyle \log_2 5(2^x) \)

16.

  1. \(\displaystyle \log_{10} \sqrt{\dfrac{2L}{R^2}} \)

  2. \(\displaystyle \log_{10} 2\pi\sqrt{\dfrac{l}{g}} \)

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. \(\displaystyle \log(x^2-x)-\log\sqrt{x^3} \)

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.

\(\log_5 (9-4x) = 3\)

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.

\(3^{2x-3} = 4529 \)

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. \(\displaystyle \log_b\dfrac{2}{5} \)

32.

  1. \(\displaystyle \log_b 75\)

  2. \(\displaystyle \log_b \sqrt{50} \)

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. \(\displaystyle (\log 2) (\log x)\)
35.
  1. \(\displaystyle \log x^3\)
  2. \(\displaystyle (\log 3)(\log x)\)
  3. \(\displaystyle 3\log x\)
  4. \(\displaystyle \log 3^x \)
36.
  1. Explain why \(~\log {\dfrac{1}{x}}~\) 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?

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 \(q\)