12.5: Solve exponential and logarithmic equations
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Even though we already discussed solving some exponential and logarithmic equations, we have yet to discuss solving exponential and logarithms equations with uncommon bases, and applying all properties. We start with a basic property of logarithms similar to the exponential equations property with common bases. Since the logarithm is one-to-one, we get the following.
Solve Logarithmic Equations Using the Properties of Logarithms
If \(a\), \(M\), \(N>0\), and \(a\neq 1\), then
\[\log_a M=\log_a N\quad\text{implies}\quad M=N\nonumber\]
Solve for \(x:\: 2\log_7x=\log_7 16\)
Solution
Since the base on both sides of the equal sign is \(7\), then we can rewrite the equation with \(\log_7\) on each side with no coefficients in front of the logarithms.
\[\begin{array}{rl} \color{blue}{2}\color{black}{\log_7}x=\log_7 16 &\text{Apply the power property of logarithms} \\ \log_7 x^{\color{blue}{2}}\color{black}{=}\log_7 16&\text{Common base, no coefficients, equate values} \\ x^2=16 &\text{Solve for }x \\ x^2-16=0&\text{Factor} \\ (x+4)(x-4)=0 &\text{Apply zero product rule} \\ x+4=0\quad\text{or}\quad x-4=0 &\text{Isolate }x \\ \cancel{x=-4 }\quad\text{or}\quad x=4 &\text{STOP Recall the domain of logarithms}\end{array}\nonumber\]
Since the domain of logarithmic functions are all values greater than zero, then we eliminate \(x = −4\) as a solution and have \(x = 4\) as the only solution to the equation. Thus, \(x = 4\) is the solution.
Solve for \(x:\:\log_4 (x+6)+\log_4 x=2\).
Solution
We will have to use the properties of logarithms, as seen in the previous section, and the definition of a logarithm to solve this problem. There are many steps, but as long as we’re organized, we will be able to obtain the solution. First, we use the product property of logarithms to rewrite the left side as a product:
\[\log_4 (x+6)+\log_4 x=\log_4((x+6)\cdot x)=\log_4 (x^2+6x)\nonumber\]
Next, we rewrite the equation using the above and the definition of a logarithm:
\[\begin{array}{rl} \log_4(x+6)+\log_4x=2 &\text{Apply the product property of logarithms} \\ \log_4(x^2+6x)=2 &\text{Rewrite in exponential form} \\ x^2+6x=4^2 & \text{Simplify }4^2 \\ x^2+6x=16 &\text{Solve for }x \\ x^2+6x-16=0 &\text{Factor} \\ (x+8)(x-2)=0 & \text{Apply the zero product rule} \\ x+8=0\quad\text{or}\quad x-2=0 &\text{Isolate }x \\ \cancel{x=-8}\quad\text{or}\quad x=2 &\text{STOP Recall the domain of logarithms}\end{array}\nonumber\]
Notice, \(x = −8\) cannot be a solution to the equation since the value of the logarithms cannot be negative. Thus, \(x = 2\) is the solution to the equation
Solve Exponential Equations
To solve exponential equations with uncommon bases, we rewrite the equations in their logarithmic form. In general, we should equate exponents when we can, but then the logarithmic form otherwise.
Solve \(2^x = 7\). Give the exact answer, and then use a calculator to approximate the exact answer to four decimal places.
Solution
When \(x\) is in the exponent, the only way to bring \(x\) down to the base position is to use the definition of a logarithm. We use this definition often when wanting to toggle between logarithmic and exponential form.
\[\begin{array}{rl}2^x =7 &\text{Uncommon bases, rewrite in logarithmic form} \\ \log_2 7=x &\text{Exact answer}\end{array}\nonumber\]
The exact answer is \(x = \log_2 7\). To approximate this value, we must use the Change of Base formula (COB):
\[\log_2 7=\dfrac{\log 7}{\log 2}\nonumber\]
Putting this in the calculator, we get \(\dfrac{\log 7}{\log 2}\approx 2.8074\). Thus, the exact answer is \(x=\log_2 7\), and the approximate answer is \(x=2.8074\).
Solve \(2e^{ x+5 }= 5\). Give the exact answer, and then use a calculator to approximate the exact answer to four decimal places.
Solution
Since we see the base of the exponential equation is \(e\), then this is a light bulb for us to use the natural logarithmic function when using the definition of a logarithm. First, we isolate the exponential equation by dividing each side by \(2\), then we rewrite the statement using the definition of a logarithm.
\[\begin{array}{rl} 2e^{x+5}=5 &\text{Divide each side by a factor }2 \\ e^{x+5}=\dfrac{5}{2}&\text{Uncommon bases, rewrite in logarithmic form} \\ \log_e\left(\dfrac{5}{2}\right)=x+5 &\text{Rewrite }\log_e\text{ as }\ln \\ \ln\left(\dfrac{5}{2}\right)=x+5 &\text{Isolate }x \\ \ln\left(\dfrac{5}{2}\right)-5=x &\text{Exact answer}\end{array}\nonumber\]
Note, \(x=\ln\left(\dfrac{5}{2}\right)-5\) is the exact solution. To approximate this value, we put this directly in the calculator. So, we get \(\ln\left(\dfrac{5}{2}\right)-5\approx -4.0837\). Thus, the exact answer is \(x=\ln\left(\dfrac{5}{2}\right)-5\), and the approximate answer is \(x = −4.0837\).
In Example 12.5.4 , we weren’t required to use the COB formula since the \(\boxed{\ln}\) is built directly into the scientific calculator. If the base is any number other than \(e\), we would have to use COB prior to putting the value into the calculator. Nowadays, some calculators have a \(\boxed{\log}\) button in which different bases other than \(10\) and \(e\) can be entered. It is just a matter of brand of calculator and identifying that feature.
We can also take the logarithm of each side of an exponential equation, as we did when developing the Change of Base Formula, to solve exponential equations.
Solve \(4^{7x} = 15\). Give the exact answer, and then use a calculator to approximate the exact answer to four decimal places.
Solution
We can take the common logarithm of each side and solve the equation.
\[\begin{array}{rl} 4^{7x}=15 &\text{Take common logarithm of each side} \\ \log 4^{7x}=\log 15 &\text{Apply power rule of logarithms} \\ 7x\log 4=\log 15&\text{Isolate }x\text{ by dividing each side by }7\log 4 \\ x=\dfrac{\log 15}{7\log 4}&\text{Exact answer}\end{array}\nonumber\]
Note, \(x = \dfrac{\log 15}{ 7 \log 4}\) is the exact solution. To approximate this value, we put this directly in the calculator. So, we get \(\dfrac{\log 15}{7\log 4}\approx 0.2971\). Thus, the exact answer is \(x=\dfrac{\log 15}{7\log 4}\), and the approximate answer is \(x = 0.2791\).
Applications with Exponential Functions
The half-life for plutonium-239 is 24,360 years. The amount \(A\) (in grams) of plutonium-239 after \(t\) years for a one-gram sample is given by \(A(t) = 1\cdot 0.5^{t/24,360}\). How long will it take before \(0.6\) gram of plutonium-\(239\) is left?
Solution
Notice the question states how long. Hence, we need to find time \(t\), for a given amount \(A\). In particularly, \(A = 0.6\). Plug-n-chug \(A = 0.6\) into the given function we get
\[\begin{array}{rl} A(t)=1\cdot 0.5^{t/24,360}&\text{Replace }A(t)=0.6 \\ 0.6=1\cdot 0.5^{t/24,360}&\text{Simplify} \\ 0.6=0.5^{t/24,360}&\text{Rewrite in logarithmic form} \\ \log_{0.5}0.6=\dfrac{t}{24,360} &\text{Isolate }t \\ t=24,360\cdot\log_{0.5}0.6 &\text{Rewrite using COB} \\ t=24,360\cdot\dfrac{\log 0.6}{\log 0.5}&\text{Exact time} \\ t\approx 17,952 &\text{Approximate time}\end{array}\nonumber\]
Thus, it will take about 17,952 years for plutonium-239 to reach \(0.6\) grams.
Solve Exponential and Logarithmic Equations Homework
Solve the equation.
\(\log_5 (x+2)-\log_5 (x-3)=3\)
\(\ln 60-\ln x=\ln (x-4)\)
\(\log_8 x=\log_8 6\)
\(\log x+\log (x+1)=\log 72\)
\(\log x+\log (x-1)=\log 72\)
\(\log (3x-8)-\log 9x=2\)
\(\log_11 (5x-6)+\log_11 x=1\)
Solve the equation. Give an exact solution and an approximate solution to four decimal places.
\(3^{x+7}=7\)
\(2^{8x}=3.6\)
\(10\cdot 2^x=11\)
\(\dfrac{1}{8}\cdot 5^{9x}=4.9\)
The half-life for thorium-227 is 18.72 days. The amount \(A\) (in grams) of thorium-239 after \(t\) years for a \(10\)-gram sample is given by
\[A(t)=10\cdot 0.5^{\dfrac{t}{18.72}}\nonumber\]
How long will it take before 4 grams of thorium-227 is left in the sample? Round your answer to the hundredths place.
According to the U.S. Census Bureau, the population of the United States in 2008 was 304 million people. In addition, the population of the United States was growing at a rate of \(1.1\%\) per year. Assuming this growth rate is continuous, the model
\[P(t)=304\cdot (1.011)^{t-2008}\nonumber\]
represents the population \(P\) (in millions of people) in year \(t\). According to the model, when will the population be 404 million people? Be sure to round your answer to the nearest whole year.
The formula \(y = 1 + 1.5 \ln(x + 1)\) models the average number of free-throws a basketball player can make consecutively during practice as a function of time, where \(x\) is the number of consecutive days the basketball player has practiced for two hours. After how many days of practice can the basketball player make an average of 8 consecutive free throws?
Newton’s Law of Cooling states that the temperature of a heated object decreases exponentially over time toward the temperature of the surrounding medium. Suppose that a coffee is served at a temperature of \(143^{\circ}\text{F}\) and placed in a room whose temperature is \(70^{\circ}\text{F}\). The temperature \(µ\) (in \(^{\circ}\text{F}\)) of the coffee at time \(t\) (in minutes) can be modeled by \(µ(t) = 70 + 73e^{−0.07t}\). When will the temperature be \(105^{\circ}\text{F}\)?