Solving Equations Using Logarithm

Section 6.4 Solving Equations Using Logarithm

This section addresses the following topics.

Read and use mathematical models in a technical document

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Communicate results in mathematical notation and in language appropriate to the technical field

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This section covers the following mathematical concepts.

Solve linear, rational, quadratic, and exponential equations and formulas (skill)

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Read and interpret models (critical thinking)

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Use models including linear, quadratic, exponential/logarithmic, and trigonometric (skill)

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This section presents methods for solving equations that have an exponential by using the relationship with the logarithm, and also how to solve some equations with logarithms.

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Subsection 6.4.1 Solving Equations with Exponentials

The first two examples demonstrate how we can use the inverse relationship between exponentials and logarithms to solve an equation with an exponential.

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Example 6.4.1.

Solve \(6.0 = 10^{\frac{3}{2}(x+10.7)}\text{.}\)

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Order of operations dictates the following

\begin{align*} 6.0 & = 10^{\frac{3}{2}(x+10.7)}.\\ \end{align*}

Use a log to undo an exponential.

\begin{align*} \log(6.0) & = \log\left(10^{\frac{3}{2}(x+10.7)}\right).\\ \log(6.0) & = \frac{3}{2}(x+10.7).\\ \frac{2}{3}\log(6.0) & = x+10.7.\\ 0.51876750 & \approx x+10.7.\\ 0.51876750-10.7 & = x.\\ -10.18123250 & \approx x.\\ -10.181 & \approx x. \end{align*}

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Rounding here was arbitrarily chosen to be 3 decimal places, because there is no context.

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Example 6.4.2.

Solve \(13 = 2+3e^{x-1}\text{.}\)

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Order of operations dictates the following

\begin{align*} 13 & = 2+3e^{x-1}.\\ 11 & = 3e^{x-1}.\\ \frac{11}{3} & = e^{x-1}\\ \end{align*}

Use a ln to undo an exponential.

\begin{align*} \ln\left(\frac{11}{3}\right) & = \ln\left(e^{x-1}\right).\\ \ln\left(\frac{11}{3}\right) & = x-1.\\ 1.29928298 & = x-1.\\ 1.29928298+1 & = x.\\ 2.29928298 & \approx x.\\ 2.299 & \approx x. \end{align*}

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Rounding here was arbitrarily chosen to be 3 decimal places, because there is no context.

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Checkpoint 6.4.3.

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The next two examples show us how to solve equations involving exponentials other than the common (base 10) or natural (base e).

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Example 6.4.4.

Solve \(5.0 = 3.0 \left(2^{\frac{t}{7\bar{0}}}\right)\text{.}\)

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Because the exponential is base 2, we will use a log base 2.

\begin{align*} 5.0 & = 3.0 \left(2^{\frac{t}{7\bar{0}}}\right).\\ 5.0/3.0 & = 2^{\frac{t}{7\bar{0}}}.\\ \log_2(5.0/3.0) & = \log_2\left(2^{\frac{t}{7\bar{0}}}\right).\\ \log_2(5.0/3.0) & = \frac{t}{7\bar{0}}.\\ 7\bar{0}\log_2(5.0/3.0) & = t. \end{align*}

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Your calculator likely does not have a button for calculating \(\log_2(x)\text{.}\) We can use a property of all logarithms to solve this equation with the natural log. In general

\begin{equation*} \log_b(a^p) = p \log_b(a)\text{.} \end{equation*}

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Example 6.4.5.

Solve \(5.0 = 3.0 \left(2^{\frac{t}{7\bar{0}}}\right)\text{.}\) Round to units.

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\begin{align*} 5.0 & = 3.0 \left(2^{\frac{t}{7\bar{0}}}\right).\\ 5.0/3.0 & = 2^{\frac{t}{7\bar{0}}}.\\ \ln(5.0/3.0) & = \ln\left(2^{\frac{t}{7\bar{0}}}\right).\\ \ln(5.0/3.0) & = \frac{t}{7\bar{0}} \cdot \ln(2).\\ \frac{7\bar{0}\ln(5.0/3.0)}{\ln(2)} & = t.\\ 51.58759159 & \approx t\\ 52 & \approx t \end{align*}

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Your calculator does have a button for \(\ln(x)\text{.}\) Note this implies that \(\log_2(5.0/3.0) = \frac{\ln(5.0/3.0)}{\ln(2)}\text{.}\) This relationship works for logs of any base.

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Checkpoint 6.4.6.

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Subsection 6.4.2 Solving Equations with Logarithms

As with equations involving exponentials, we can solve equations involving logarithms using the inverse relationship between exponentials and logarithms. The first two examples demonstrate using the definition of logarithm.

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Example 6.4.7.

Solve

\begin{equation*} \log_3(x)=2\text{.} \end{equation*}

We can re-write this as

\begin{equation*} 3^2=x \end{equation*}

which tells us that \(x=9\text{.}\)

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Example 6.4.8.

Solve \(14 = 12+\log(5x)\text{.}\)

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First we combine the terms outside the logarithm, the we re-write that as an exponential.

\begin{align*} 14 & = 12+\log(5x)\\ 2 & = \log(5x)\\ \end{align*}

Re-write the logarithm as an exponential.

\begin{align*} 10^2 & = 5x\\ 100 & = 5x\\ 20 & = x \end{align*}

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Example 6.4.9.

Solve \(7 = \ln(3x+2)\text{.}\) Round to units.

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Solution.

We re-write this as an exponential.

\begin{align*} 7 & = \ln(3x+2)\\ e^7 & = 3x+2\\ e^7-2 & = 3x\\ \frac{1}{3}(e^7-2) & = x\\ 365 & \approx x \end{align*}

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Checkpoint 6.4.10.

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Subsection 6.4.3 Applications with Exponentials and Logarithms

Example 6.4.11.

In Example 6.1.13 we produced the model \(P=3.0 \cdot 2^{t/7\bar{0}}\text{.}\) Here we will redo this problem using the model \(P=P_0 e^{rt}\text{.}\)

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The bacteria lactobacilus acidophilus doubles in population every 70 minutes. If the initial amount was 3 grams, what how much would there be after 24 hours?

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To use this model we must first calculate the constant \(k\text{.}\) We know the amount doubles every 70 minutes, so we can start with the following.

\begin{align*} 6.0 & = 3.0e^{7\bar{0} r}.\\ \frac{6.0}{3.0} & = e^{7\bar{0}r}.\\ 2.0 & = e^{7\bar{0}r}.\\ \ln(2.0) & = \ln\left(e^{7\bar{0}r}\right).\\ \ln(2.0) & = 7\bar{0}r.\\ \frac{\ln(2.0)}{7\bar{0}} & = r.\\ k & \approx 0.009902.\\ & \approx 0.0099\text{.} \end{align*}

Using this value we can now calculate the value after 24 hours. Note 24 hours is \(24 \cdot 60 = 1440\) minutes.

\begin{align*} P & = 3.0e^{(144\bar{0})(0.0099)}\\ & \approx 4.674533843 \times 10^6\\ & \approx 4.7 \times 10^6 \end{align*}

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This is \(4.7 \times 10^3 \approx 47,000\) kilograms. Naturally, this is an unreasonable prediction. This tells us there must be other factors in bacteria growth.

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Example 6.4.12.

Plutonium-241 has a half-life of 14.4 years. This means if you start with 10.0 g of Pu-241 in 14.4 years there will be only 5.0 g of Pu-241. Generally, this can also be modeled by

\begin{equation*} P=P_0 e^{rt}\text{.} \end{equation*}

\(P_0\) is the initial amount of material. \(P\) is the amount left after \(t\) units of time. \(r\) is a constant that is derives from the speed how fast the material decays.

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(a)

Write the model for Plutonium-241.

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Solution.

We must first calculate \(k\text{.}\) We can use Example 6.4.1 as an example.

\begin{align*} 5.0 & = 10.0 e^{r(14.4)}\\ \frac{1}{2.0} & = e^{r(14.4)}\\ \ln\left(\frac{1}{2.0}\right) & = \ln\left(e^{r(14.4)}\right)\\ \ln\left(\frac{1}{2.0}\right) & = r(14.4).\\ \frac{1}{14.4}\ln\left(\frac{1}{2.0}\right) & = r\\ -0.04813522 & \approx r.\\ -0.048 & \approx r. \end{align*}

Thus the model is

\begin{equation*} P = P_0 e^{-0.048t}\text{.} \end{equation*}

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(b)

If a lab has 12.0g of Pu-241, how much will be left in 6 years?

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Solution.

We use the model from the previous step.

\begin{align*} P & = 12.0e^{-0.048(6)}\\ & \approx 8.997139\\ & \approx 9.0. \end{align*}

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Note, that the example in Example 6.1.13 and Example 6.4.11 imply that \(3.0 \cdot 2^{t/70} = 3.0 e^{(0.0099)t}\text{.}\) In particular we can convert the power of 2 to a power of e. More generally, we can write \(2^x = e^{rx}\) or \(3^x = e^{rx}\) or similar for some value of \(k\text{.}\) The next example shows how we can perform this conversion.

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Example 6.4.13.

Write \(2^x\) as \(e^{rx}\text{.}\) Numbers are exact.

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\begin{align*} 2^x & = e^{rx}.\\ \ln(2^x) & = \ln(e^{rx}).\\ x \ln(2) & = (rx)\ln(e).\\ x \ln(2) & = rx.\\ \frac{x \ln(2)}{x} & = \frac{rx}{x}.\\ \ln(2) & = r \end{align*}

Thus \(2^x = e^{x \ln(2)}\text{.}\)

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Definition 6.4.14. pH (percent hydrogen).

Acidity is measured in pH (percent hydrogen). The calculation is

\begin{equation*} \text{pH} = -\log(H_3O^+) \end{equation*}

where \(H_3O^+\) is the concentration of hydronium ions per mole. This is obtained experimentally.

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Example 6.4.15.

(a)

A solution of hydrochloric acid has a concentration of 0.0025. Calculate its pH.

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\begin{align*} \text{pH} & = -\log(0.0025)\\ & \approx 2.6. \end{align*}

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(b)

Sweat has a pH between 4.5 and 7. Suppose sweat is measured to have a pH of 5.3. Determine the concentration of ions.

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We setup the pH calculation and solve using Example 6.4.9.

\begin{align*} 5.3 & = -\log(c)\\ -5.3 & = \log(c)\\ 10^{-5.3} & = 10^{\log(c)}\\ 10^{-5.3} & = c\\ 5.0 \times 10^{-6} & \approx c \end{align*}

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What is the purpose behind defining pH using a log? What does it do, that simply giving the concentration of \(H_3O^+\) does not? The next example illustrates what the use of a log adds.

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Example 6.4.16.

(a)

Calculate the concentration of \(H_3O^+\) for pH levels of 5, 6, 7 (all within the range of human sweat).

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Solution.

\begin{align*} 5 & = -\log(c_5).\\ 10^{-5} & = c_5.\\ 6 & = -\log(c_6).\\ 10^{-6} & = c_6.\\ 7 & = -\log(c_7).\\ 10^{-7} & = c_7. \end{align*}

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(b)

What is the ratio of the concentration from pH of 5 to 6? of 6 to 7? of 5 to 7? This means calculate the ratios \(c_5/c_6\text{,}\) \(c_6/c_7\text{,}\) and \(c_5/c_7\text{.}\)

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Solution.

\begin{align*} \frac{c_5}{c_6} & =\frac{10^{-5}}{10^{-6}} = 10.\\ \frac{c_6}{c_7} & =\frac{10^{-6}}{10^{-7}} = 10.\\ \frac{c_5}{c_7} & =\frac{10^{-5}}{10^{-7}} = 100. \end{align*}

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We see that a change of one pH means the substance is 10 times as strong. A change of 2 pH means the substance is \(10^2=100\) times as strong. The log scale gives us growth as a ratio.

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Definition 6.4.17. Moment Magnitude.

Larger earthquakes today are measured and reported using the moment magnitude scale. This is calculated via

\begin{equation*} M_w = \frac{2}{3}\log(M_0)-10.7 \end{equation*}

where \(M_w\) is the moment magnitude, and \(M_0\) is the seismic moment in Newtons per meter (a measure of energy).

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Example 6.4.18.

(a)

Based on seismic readings \(M_0 = 7.2 \times 10^{22}\text{.}\) What was the moment magnitude? These are rounded to one decimal place.

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Solution.

Using the formula we obtain

\begin{align*} M_w & = \frac{2}{3}\log(7.2 \times 10^{22})-10.7\\ & \approx 4.53822167\\ & \approx 4.5 \end{align*}

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(b)

What was the seismic moment for an earthquake with magnitude 7.1?

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Solution.

We setup the calculation and solve using Example 6.4.9.

\begin{align*} 7.1 & = \frac{2}{3}\log(M_0)-10.7\\ 17.8 & = \frac{2}{3}\log(M_0)\\ 26.7 & = \log(M_0)\\ 10^{26.7} & = M_0\\ 5.011872336 \times 10^{26} & \approx M_0\\ 5.01 \times 10^{26} & \approx M_0 \end{align*}

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Just as with pH, the moment magnitude seismic scale enables us to compare how much stronger one earthquake is than another. The 2/3 means that a change of 1 is not 10 times as strong but a different ratio. Can you figure out what that ratio is?

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