Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

8.5: Solving Exponentials for Time- Logarithms

  • Page ID
    34222
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    Earlier, we found that since Olympia, WA had a population of 245 thousand in 2008 and had been growing at 3% per year, the population could be modeled by the equation

    \[P_{n}=(1+0.03)^{n}(245,000) \nonumber\]

    or equivalently,

    \[P_{n}=245,000(1.03)^{n}. \nonumber\]

    Using this equation, we were able to predict the population in the future.

    Suppose we wanted to know when the population of Olympia would reach 400 thousand. Since we are looking for the year n when the population will be 400 thousand, we would need to solve the equation

    \begin{aligned}
    &400,000=245,000(1.03)^{n} \quad \text { dividing both sides by } 245,000 \text { gives }\\
    &1.6327=1.03^{n}
    \end{aligned}

    One approach to this problem would be to create a table of values, or to use technology to draw a graph to estimate the solution.

    A graph of 245 times 1.03 to the n, which gently curves upwards.  When the input is 16 the output is about 390, and when the input is 17 the output is about 410.

    From the graph, we can estimate that the solution will be around 16 to 17 years after 2008 (2024 to 2025). This is pretty good, but we’d really like to have an algebraic tool to answer this question. To do that, we need to introduce a new function that will undo exponentials, similar to how a square root undoes a square. For exponentials, the function we need is called a logarithm. It is the inverse of the exponential, meaning it undoes the exponential. While there is a whole family of logarithms with different bases, we will focus on the common log, which is based on the exponential 10x.

    Common Logarithm

    The common logarithm, written \(\log(x)\), undoes the exponential \(10^x\)

    This means that

    \[\log(10^x) = x\]

    and likewise

    \[10^{\log(x)} = x\]

    This also means the statement \(10^a = b\) is equivalent to the statement \(\log(b) = a\)

    \(\log(x)\) is read as “\(\log\) of \(x\)”, and means “the logarithm of the value \(x\)”. It is important to note that this is not multiplication – the log doesn’t mean anything by itself, just like √ doesn’t mean anything by itself; it has to be applied to a number.

    Example 9

    Evaluate each of the following

    1. \(\log(100)\)
    2. \(\log(1000)\)
    3. \(\log(10000)\)
    4. \(\log(1/100)\)
    5. \(\log(1)\)

    Solution

    1. \(\log(100)\) can be written as \(\log(10^2)\). Since the log undoes the exponential, \(\log(10^2) = 2\)
    2. \(\log(1000) = \log(10^3) = 3\)
    3. \(\log(10000) = \log(10^4) = 4\)
    4. Recall that \(x^{-n}=\frac{1}{x^{n}}\). \(\log \left(\frac{1}{100}\right)=\log \left(10^{-2}\right)=-2\)
    5. Recall that \(x^0 = 1\). \(\log(1) = \log(10^0) = 0\)

    It is helpful to note that from the first three parts of the previous example that the number we’re taking the log of has to get \(10\) times bigger for the log to increase in value by 1.

    Of course, most numbers cannot be written as a nice simple power of 10. For those numbers, we can evaluate the log using a scientific calculator with a log button.

    Example 10

    Evaluate \(\log(300)\)

    Solution

    Using a calculator, \(\log(300)\) is approximately \(2.477121\)

    With an equation, just like we can add a number to both sides, multiply both sides by a number, or square both sides, we can also take the logarithm of both sides of the equation and end up with an equivalent equation. This will allow us to solve some simple equations.

    Example 11

    1. Solve \(10^x = 1000\)
    2. Solve \(10^x = 3\)
    3. Solve \(2(10^x) = 8\)

    Solution

    a) Taking the log of both sides gives \(\log(10^x) = \log(1000)\)

    Since the log undoes the exponential, \(\log(10^x) = x\). Similarly \(\log(1000) = \log(10^3) = 3\).

    The equation simplifies then to \(x = 3\).

    b) Taking the log of both sides gives \(\log(10^x) = \log(3)\).

    On the left side, \(\log(10^x) = x\), so \(x = \log(3)\). We can approximate this value with a calculator. \(x \approx 0.477\)

    c) Here we would first want to isolate the exponential by dividing both sides of the equation by 2, giving

    \(10^x = 4\).

    Now we can take the log of both sides, giving \(\log(10^x) = \log(4)\), which simplifies to

    \(x = \log(4) \approx 0.602\)

    This approach allows us to solve exponential equations with powers of 10, but what about problems like \(2 = 1.03^n\) from earlier, which have a base of 1.03? For that, we need the exponent property for logs.

    Properties of Logs: Exponent Property

    \(\log \left(A^{r}\right)=r \log (A)\)

    To show why this is true, we offer a proof.

    Since the logarithm and exponential undo each other, \(10^{\log A}=A\). So

    \[A^{r}=\left(10^{\log \cdot A}\right)^{r} \nonumber\]

    Utilizing the exponential rule that states \(\left(x^{a}\right)^{b}=x^{a b}\),

    \[A^{r}=\left(10^{\log A}\right)^{r}=10^{r \log A} \nonumber\]

    So then

    \[\log \left(A^{r}\right)=\log \left(10^{r \log A}\right) \nonumber\]

    Again utilizing the property that the log undoes the exponential on the right side yields the result

    \[\log \left(A^{r}\right)=r \log A \nonumber\]

    Example 12

    Rewrite log(25) using the exponent property for logs

    Solution

    \[\log (25)=\log \left(5^{2}\right)=2 \log (5) \nonumber\]

    This property will finally allow us to answer our original question.

    Solving exponential equations with logarithms

    1. Isolate the exponential. In other words, get it by itself on one side of the equation. This usually involves dividing by a number multiplying it.
    2. Take the log of both sides of the equation.
    3. Use the exponent property of logs to rewrite the exponential with the variable exponent multiplying the logarithm.
    4. Divide as needed to solve for the variable.

    Example 13

    If Olympia is growing according to the equation, \(P_{n}=245(1.03)^{n}\), where n is years after 2008, and the population is measured in thousands. Find when the population will be 400 thousand.

    Solution

    We need to solve the equation

    \(\begin{array}{ll} 400=245(1.03)^{n} & \text{Begin by dividing both sides by 245 to isolate the exponential} \\
    1.633=1.03^{n} & \text{Now take the log of both sides} \\ \log (1.633)=\log \left(1.03^{n}\right) & \text{Use the exponent property of logs on the right side} \\ \log (1.633)=n \log (1.03) & {\text{Now we can divide by} \log (1.03)} \\
    \frac{\log (1.633)}{\log (1.03)}=n &
    \text{We can approximate this value on a calculator} \\
    n \approx 16.591^{\circ} \end{array}\)

    Alternatively, after applying the exponent property of logs on the right side, we could have evaluated the logarithms to decimal approximations and completed our calculations using those approximations, as you’ll see in the next example. While the final answer may come out slightly differently, as long as we keep enough significant values during calculation, our answer will be close enough for most purposes.

    Example 14

    Polluted water is passed through a series of filters. Each filter removes 90% of the remaining impurities from the water. If you have 10 million particles of pollutant per gallon originally, how many filters would the water need to be passed through to reduce the pollutant to 500 particles per gallon?

    Solution

    In this problem, our “population” is the number of particles of pollutant per gallon. The initial pollutant is 10 million particles per gallon, so \(P_0 = 10,000,000\). Instead of changing with time, the pollutant changes with the number of filters, so \(n\) will represent the number of filters the water passes through.

    Also, since the amount of pollutant is decreasing with each filter instead of increasing, our “growth” rate will be negative, indicating that the population is decreasing instead of increasing, so \(r = -0.90\).

    We can then write the explicit equation for the pollutant:

    \[\begin{algin*} P_{n} &=10,000,000(1-0.90)^{n} \\[4pt] &=10,000,000(0.10)^{n} \end{align*}\]

    To solve the question of how many filters are needed to lower the pollutant to 500 particles per gallon, we can set \(P_n\) equal to 500, and solve for \(n\).

    \(\begin{array}{ll} 500=10,000,000(0.10)^{n} & \text{Divide both sides by} 10,000,000 \\
    0.00005=0.10^{n} & \text{Take the log of both sides} \\ \log (0.00005)=\log \left(0.10^{n}\right) &
    \text{Use the exponent property of logs on the right side} \\ \log (0.00005)=n \log (0.10) &
    \text{Evaluate the logarithms to a decimal approximation} \\ -4.301=n(-1) & \text{Divide by} -1,\text{ the value multiplying }n \\ 4.301=n \end{array}\)

    It would take about 4.301 filters. Of course, since we probably can’t install 0.3 filters, we would need to use 5 filters to bring the pollutant below the desired level.

    Try it Now 4

    India had a population in 2008 of about 1.14 billion people. The population is growing by about 1.34% each year. If this trend continues, when will India’s population reach 1.2 billion?

    Answer

    \(1.14(1.0134)^{n}=1.2\). \(n = 3.853\), which is during 2011


    8.5: Solving Exponentials for Time- Logarithms is shared under a CC BY-SA 3.0 license and was authored, remixed, and/or curated by David Lippman via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.