MTH120 College Algebra Chapter 6.6

 6.6 Exponential and Logarithmic Equations

Exponential and logarithmic equations involve equations with exponential or logarithmic expressions. These types of equations are solved by applying the properties and rules of exponentials and logarithms. Here are some key strategies and steps for solving exponential and logarithmic equations:

Exponential Equations:

1. Isolate the exponential term: If you have an equation like ${�}^{�}=�$, where $�$ and $�$ are constants, isolate ${�}^{�}$ on one side of the equation.

2. Take the logarithm of both sides: Use a logarithm with a base that simplifies the equation. Common choices include natural logarithm ($��$) or common logarithm ($���$). Apply the logarithm to both sides to bring down the exponent.

3. Apply logarithmic properties: Use logarithmic properties to simplify the equation, often changing the equation from an exponential form to a linear form.

4. Solve for the variable: After simplifying, solve for the variable. Be sure to consider any possible extraneous solutions.

Logarithmic Equations:

1. Isolate the logarithmic term: If you have an equation like $��{�}_{�}(�)=�$, where $�$ and $�$ are constants, isolate the logarithmic term on one side of the equation.

2. Apply the inverse operation: The inverse operation of a logarithm is exponentiation. Use exponentiation with the same base as the logarithm to remove the logarithmic term.

3. Solve for the variable: After removing the logarithmic term, solve for the variable.

Here are some common examples of exponential and logarithmic equations:

Exponential Equation: $2^{�}=16$

Steps to Solve:

1. Isolate the exponential term: $2^{�}=16$
2. Take the logarithm of both sides: $�\cdot {\log }_2(2)={\log }_2(16)$
3. Simplify using logarithmic properties: $�=4$

Logarithmic Equation: $��{�}_{10}(�)=2$

Steps to Solve:

1. Isolate the logarithmic term: $��{�}_{10}(�)=2$
2. Apply the inverse operation: $�=10^2$
3. Solve for the variable: $�=100$

These are basic examples, and more complex equations may require additional steps and considerations. When solving exponential and logarithmic equations, always verify your solutions and check for possible extraneous solutions.

To solve exponential equations using like bases, you need to make sure that both sides of the equation have the same base. This allows you to set the exponents equal to each other. Here are the steps to solve exponential equations with like bases:

Step 1: Ensure both sides have the same base.

Make sure that both sides of the equation have the same base. If they don't, rewrite one side of the equation so that the bases match.

Step 2: Set the exponents equal to each other.

Once both sides have the same base, set the exponents equal to each other. The equation should look like this:

${�}^{�}={�}^{�}$

Step 3: Solve for the variable.

Since the bases are the same, you can set the exponents equal to each other:

$�=�$

Now, you have solved the equation and found the value of the variable.

Example: Let's solve the equation $2^{�}=2^4$.

Step 1: Both sides have the same base, which is 2.

Step 2: Set the exponents equal to each other:

$�=4$

Step 3: Solve for the variable:

$�=4$

So, the solution to the equation $2^{�}=2^4$ is $�=4$.

This method works when you have exponential equations with the same base on both sides. It simplifies the process and allows you to solve for the variable directly.

To solve an exponential equation of the form ${�}^{�}={�}^{�}$, where $�$ and $�$ are algebraic expressions with an unknown variable, you can use the property of logarithms that states if two expressions with the same base are equal, then their exponents must be equal. Here are the steps to solve for the unknown variable:

1. Write the equation: Start with the exponential equation ${�}^{�}={�}^{�}$, where $�$ and $�$ are algebraic expressions with the unknown variable.

2. Apply the logarithm: Take the logarithm of both sides of the equation. You can use any base for the logarithm (common logarithm, natural logarithm, or any other base) as long as it's consistent on both sides. Let's use the natural logarithm $\ln$ for this example:

   $\ln ({�}^{�})=\ln ({�}^{�})$

3. Apply the power rule of logarithms: The power rule of logarithms allows you to bring the exponents down as coefficients:

   $�\cdot \ln (�)=�\cdot \ln (�)$

4. Divide by $\ln (�)$: To solve for the unknown variable, divide both sides by $\ln (�)$:

   $�=�$

Now you have solved for the unknown variable, and the solution is $�=�$.

This method works when you have an exponential equation with the same base on both sides and algebraic expressions for the exponents. By taking the logarithm of both sides, you can simplify the equation and solve for the unknown variable.

To solve equations with exponential terms that have different bases, you can rewrite the equation so that all the powers have the same base. This allows you to set the exponents equal to each other and solve for the variable. Here's how you can rewrite equations with different bases to have a common base:

Step 1: Identify the bases. Look at the different bases in the equation. If you have bases like $�$, $�$, and $�$, choose a common base that can be applied to all terms. Often, it's helpful to choose a base that simplifies the equation.

Step 2: Rewrite the terms. Rewrite each term with the common base you've chosen. You may need to use the properties of exponents to rewrite the terms. The idea is to express all terms with the same base.

Step 3: Set the exponents equal to each other. Once all the terms have the same base, set the exponents equal to each other:

${�}^{�}={�}^{�}={�}^{�}$

Step 4: Solve for the variable. With the exponents equal, you can now solve for the variable, whether it's $�$, $�$, or $�$.

Example:

Let's say you have the equation $2^{�}=4^3$. Here's how you can rewrite it to have a common base:

Step 1: Identify the bases. In this case, the bases are 2 and 4.

Step 2: Rewrite the terms. We can rewrite 4 as $2^2$:

$2^{�}=(2^2{)}^3$

Now all terms have the base 2.

Step 3: Set the exponents equal to each other:

$�=2\cdot 3$

Step 4: Solve for the variable:

$�=6$

So, the solution to the equation $2^{�}=4^3$ is $�=6$.

By choosing a common base and rewriting the terms, you can simplify equations with different bases and solve for the variable.

Solving an equation with both positive and negative powers may involve algebraic manipulation and the use of properties of exponents. The goal is to isolate the variable and solve for its value. Here are the steps to solve an equation with positive and negative powers:

Step 1: Identify the terms with positive and negative powers.

Look for terms in the equation with both positive and negative exponents. In this context, let's consider the variable as $�$ and the equation may have terms like ${�}^{�}$ and ${�}^{-�}$.

Step 2: Move terms with negative exponents to the other side.

If you have terms with negative exponents on one side of the equation, move them to the other side. This involves changing the sign of the exponent when the term crosses the equal sign.

Step 3: Combine like terms.

Once you have moved the terms with negative exponents to the other side, try to simplify the equation by combining like terms. For example, if you have ${�}^{�}$ on one side and ${�}^{-�}$ on the other side, these terms cancel out.

Step 4: Isolate the variable.

After combining like terms, you should have an equation with the variable isolated on one side. The equation may look something like ${�}^{�}=�$, where $�$ is a constant.

Step 5: Solve for the variable.

Solve the equation for the variable $�$ using the appropriate method, depending on the form of the equation. This could involve taking the appropriate root or applying logarithms, depending on the specific equation.

Example:

Let's say you have the equation ${�}^3-{�}^{-2}=7$. Here's how you can solve it:

Step 1: Identify the terms with positive and negative powers. In this equation, ${�}^3$ has a positive exponent, and ${�}^{-2}$ has a negative exponent.

Step 2: Move the term with the negative exponent to the other side:

${�}^3=7+{�}^{-2}$

Step 3: Combine like terms. Since ${�}^{-2}$ and ${�}^3$ are on opposite sides of the equation, they don't combine directly.

Step 4: Isolate the variable. We have ${�}^3$ isolated on the left side:

${�}^3=7+{�}^{-2}$

Step 5: Solve for the variable. To solve for $�$, you can use a method appropriate to the equation's form. In this case, you can subtract ${�}^{-2}$ from both sides and then take the cube root of both sides:

$�=\sqrt[3]{7+{�}^{-2}}$

Solving equations with both positive and negative powers may require various techniques, depending on the specific equation. In some cases, it may involve taking roots, applying logarithms, or other algebraic methods to isolate and solve for the variable.

Solving exponential equations using logarithms is a common technique when the variable is in the exponent. To solve such equations, you typically take the logarithm of both sides to bring the exponent down and then solve for the variable. Here are the steps to solve exponential equations using logarithms:

Step 1: Identify the exponential equation. You should have an equation in the form ${�}^{�}=�$, where $�$ and $�$ are constants.

Step 2: Take the logarithm of both sides. Choose a logarithm base (common logarithm, natural logarithm, or any other base) and apply it to both sides of the equation. For this example, let's use the natural logarithm $\ln$:

${�}^{�}=�\Rightarrow \ln ({�}^{�})=\ln (�)$

Step 3: Apply the logarithm properties. Use logarithmic properties to simplify the equation. The exponent can be brought down as a coefficient using the power rule of logarithms:

$�\cdot \ln (�)=\ln (�)$

Step 4: Solve for the variable. Now that the variable is no longer in the exponent, you can solve for it by dividing by $\ln (�)$:

$�=\frac{\ln (�)}{\ln (�)}$

Step 5: Use a calculator. If needed, use a calculator to compute the numerical value of $�$ by evaluating the natural logarithms.

Example:

Let's solve the exponential equation $3^{�}=27$:

Step 1: Identify the exponential equation: $3^{�}=27$.

Step 2: Take the natural logarithm of both sides:

$\ln (3^{�})=\ln (27)$.

Step 3: Apply the logarithm properties:

$�\cdot \ln (3)=\ln (27)$.

Step 4: Solve for $�$:

$�=\frac{\ln (27)}{\ln (3)}$.

Step 5: Use a calculator to evaluate:

$�\approx 3$.

So, the solution to the equation $3^{�}=27$ is $�\approx 3$.

This method is applicable to exponential equations where the variable is in the exponent and allows you to find the value of the variable using logarithmic properties and a calculator if necessary.

Equations containing the mathematical constant $�$ often involve exponential growth or decay. To solve equations with $�$, you'll typically use natural logarithms ($\ln$), which is the inverse of the exponential function with base $�$. Here are the steps to solve equations containing $�$:

Step 1: Identify the equation. You should have an equation that contains $�$, such as ${�}^{�}=�$, where $�$ is a constant.

Step 2: Take the natural logarithm of both sides. Apply the natural logarithm ($\ln$) to both sides of the equation:

${�}^{�}=�\Rightarrow \ln ({�}^{�})=\ln (�)$

Step 3: Apply the logarithm properties. The exponent $�$ can be brought down as a coefficient using the power rule of logarithms:

$�\cdot \ln (�)=\ln (�)$

Since $\ln (�)=1$, this simplifies to:

$�=\ln (�)$

Step 4: Solve for the variable. Solve for $�$ by isolating it on one side of the equation:

$�=\ln (�)$

Step 5: Use a calculator. If necessary, use a calculator to compute the value of $�$ by evaluating the natural logarithm.

Example:

Let's solve the equation ${�}^{�}=20$:

Step 1: Identify the equation: ${�}^{�}=20$.

Step 2: Take the natural logarithm of both sides:

$\ln ({�}^{�})=\ln (20)$.

Step 3: Apply the logarithm properties:

$�\cdot \ln (�)=\ln (20)$.

Since $\ln (�)=1$, this simplifies to:

$�=\ln (20)$

Step 4: Solve for (x:

$�=\ln (20)$

Step 5: Use a calculator to evaluate:

$�\approx 2.9957$

So, the solution to the equation ${�}^{�}=20$ is approximately $�\approx 2.9957$.

This method allows you to solve equations containing the constant $�$ by using natural logarithms to isolate and solve for the variable $�$.

Extraneous solutions are solutions that appear to satisfy an equation but, upon further examination, do not actually work in the original problem. These can arise when solving equations, especially equations involving radicals or logarithms. Extraneous solutions occur because certain operations, like taking the square root or the logarithm, introduce restrictions or conditions on the domain of the equation.

Here are some key points to keep in mind when dealing with extraneous solutions:

1. Always check your solutions: When you solve equations, particularly those involving radicals or logarithms, it's essential to check your solutions in the original equation to ensure they are valid.

2. Identify potential issues: Pay attention to operations that can introduce restrictions. For example, when taking the square root of both sides, you must consider the possibility of both positive and negative roots. Similarly, when taking the logarithm, consider the domain of the logarithmic function.

3. Reject extraneous solutions: If a solution does not satisfy the original equation or falls outside the domain of the problem, it is considered extraneous and should be rejected.

Examples:

1. Quadratic equation: Consider the equation ${�}^2=4$. When taking the square root of both sides, you obtain $�=\pm 2$. However, if the original problem was, for instance, "Find the length of a side of a square," negative values may not be relevant in the context of the problem, so the solution $�=-2$ could be extraneous.

2. Logarithmic equation: Solve the equation $\ln (�)=1$. Taking the exponential of both sides gives $�={�}^1=�$. However, the natural logarithm is only defined for positive values of $�$, so $�=�$ is the valid solution, and $�=-�$ is an extraneous solution.

3. Radical equation: Solve the equation $\sqrt{�+3}=2$. Squaring both sides gives $�+3=4$, which leads to $�=1$. However, you must check the original equation to ensure it's valid. If the original problem states that $�$ represents the number of items in a collection, and the collection can't have negative quantities, then $�=1$ is the valid solution, and $�=-3$ is extraneous.

In summary, always be cautious when solving equations and double-check your solutions to make sure they are valid in the context of the problem. Extraneous solutions can lead to incorrect or irrelevant results if not identified and rejected.

Solving logarithmic equations using the definition of a logarithm involves expressing the equation in a way that allows you to isolate the variable. To do this, you can use the definition of a logarithm to convert between logarithmic form and exponential form, depending on the structure of the equation. Here are the steps to solve logarithmic equations using the definition of a logarithm:

Step 1: Identify the logarithmic equation. You should have an equation in the form ${\log }_{�}(�)=�$, where $�$ is the base of the logarithm, $�$ is the variable, and $�$ is a constant.

Step 2: Express the equation in exponential form. Use the definition of a logarithm to convert the equation to exponential form. For a logarithm with base $�$, the exponential form is ${�}^{�}=�$. This means that $�$ raised to the power of $�$ equals $�$.

Step 3: Solve for the variable. With the equation in exponential form, you can directly solve for the variable $�$ by finding the base $�$ raised to the power $�$.

Example:

Let's solve the logarithmic equation ${\log }_2(�)=3$:

Step 1: Identify the logarithmic equation: ${\log }_2(�)=3$.

Step 2: Express the equation in exponential form using the definition of a logarithm: $2^3=�$.

Step 3: Solve for the variable:

$2^3=8$

So, $�=8$.

The solution to the logarithmic equation ${\log }_2(�)=3$ is $�=8$.

This method allows you to solve logarithmic equations by converting them to exponential form and directly solving for the variable using the definition of a logarithm.

The one-to-one property of logarithms is a powerful tool for solving logarithmic equations. It states that for any positive real numbers $�$, $�$, and $�$, if ${\log }_{�}(�)={\log }_{�}(�)$, then $�=�$. In other words, if the logarithms have the same base and are equal, the arguments of the logarithms must also be equal. Here are the steps to solve logarithmic equations using the one-to-one property:

Step 1: Identify the logarithmic equation. You should have an equation in the form ${\log }_{�}(�)={\log }_{�}(�)$, where $�$ is the base of the logarithm, and $�$ and $�$ are expressions.

Step 2: Apply the one-to-one property. Since the logarithms have the same base, if ${\log }_{�}(�)={\log }_{�}(�)$, then $�=�$. This allows you to equate the arguments of the logarithms.

Step 3: Solve for the variable. With $�=�$, you can now solve for the variable.

Example:

Let's solve the logarithmic equation ${\log }_5(2�+3)={\log }_5(3�-1)$:

Step 1: Identify the logarithmic equation: ${\log }_5(2�+3)={\log }_5(3�-1)$.

Step 2: Apply the one-to-one property to equate the arguments of the logarithms:

$2�+3=3�-1$.

Step 3: Solve for the variable:

First, move the $2�$ term to the other side of the equation:

$3�-2�=3+1$.

$�=4$.

The solution to the logarithmic equation ${\log }_5(2�+3)={\log }_5(3�-1)$ is $�=4$.

This method is especially useful for solving logarithmic equations with the same base, and it simplifies the process by allowing you to directly equate the arguments of the logarithms.

Solving applied problems using exponential and logarithmic equations involves translating real-world situations into mathematical equations, solving those equations, and then interpreting the solutions in the context of the problem. Here are the steps to solve applied problems using exponential and logarithmic equations:

Step 1: Understand the problem. Carefully read and understand the problem statement, identifying key information, variables, and what needs to be found.

Step 2: Model the situation. Translate the problem into a mathematical model, typically using exponential or logarithmic equations. Determine which equations and functions best represent the problem.

Step 3: Solve the equations. Use the appropriate mathematical techniques to solve the equations. This may involve algebra, logarithms, exponents, or a combination of these.

Step 4: Interpret the solutions. Once you have solved the equations, interpret the solutions in the context of the problem. Ensure that your solutions make sense and address the original question or requirement.

Step 5: Check your work. Double-check your solutions to verify their correctness and that they are consistent with the problem's requirements.

Step 6: Communicate the results. Present your solutions in a clear and organized manner, including any relevant units or measurements. Make sure your answer addresses the original problem.

Example:

Let's solve an applied problem using exponential and logarithmic equations:

Problem: The population of a city is growing exponentially at a rate of 3% per year. If the current population is 10,000, find the population of the city after 5 years.

Step 1: Understand the problem. We have a population growth problem, and we want to find the population after a specific time period.

Step 2: Model the situation. We can use the formula for exponential growth:

$�(�)={�}_0\cdot {�}^{��}$

where:

• $�(�)$ is the population after time $�$.
• ${�}_0$ is the initial population (10,000 in this case).
• $�$ is the annual growth rate (3% or 0.03 as a decimal).
• $�$ is the time in years (5 years in this case).

Step 3: Solve the equations. Substituting the values into the formula, we get:

$�(5)=10,000\cdot {�}^{0.03\cdot 5}$

Using a calculator to compute this, we find $�(5)\approx 11,591.33$.

Step 4: Interpret the solutions. After 5 years, the population of the city is approximately 11,591.33.

Step 5: Check your work. Make sure that the result is reasonable. In this case, it is, as the population is expected to grow over time with a positive growth rate.

Step 6: Communicate the results. Present the answer clearly: After 5 years, the population of the city is approximately 11,591.33.

Solving applied problems using exponential and logarithmic equations allows you to address real-world scenarios, make predictions, and analyze various situations by applying mathematical models and problem-solving skills.

For the following exercises, solve each equation for x. 1. log(x+16)=log(x)+log(18) 2. ln(x)+ln(x−6)=ln(4x) 3. log2(3x+7)=2 4. ln(5)+7n(1−2x1)=ln(18) 5. log6(x+9)−log4(x)=log4(49) 6. ln(2)−ln(2−2x)=ln(6) 7. log2(1x)−log2(5)=log2(87)

Let's solve each of these logarithmic and natural logarithmic equations for $�$:

1. $\log (�+16)=\log (�)+\log (18)$

Using the properties of logarithms, we can combine the right side into a single logarithm:

$\log (�+16)=\log (18�)$

Now, both sides have the same base, which is 10. According to the one-to-one property of logarithms, if ${\log }_{�}(�)={\log }_{�}(�)$, then $�=�$.

So, $�+16=18�$.

Now, solve for $�$:

$16=18�-�$

$16=17�$

$�=\frac{16}{17}$

2. $\ln (�)+\ln (�-6)=\ln (4�)$

Using the properties of natural logarithms, we can combine the left side into a single logarithm:

$\ln (�(�-6))=\ln (4�)$

Now, both sides have the same base, which is $�$ (the base of natural logarithms). According to the one-to-one property of logarithms, if $\ln (�)=\ln (�)$, then $�=�$.

So, $�(�-6)=4�$.

Now, solve for (x:

${�}^2-6�=4�$

${�}^2-6�-4�=0$

${�}^2-10�=0$

$�(�-10)=0$

From here, we have two potential solutions:

$�=0$ or $�-10=0\Rightarrow �=10$

3. ${\log }_2(3�+7)=2$

Using the properties of logarithms, we can rewrite this equation in exponential form:

$3�+7=2^2$

Simplify:

$3�+7=4$

Now, solve for (x:

$3�=4-7$

$3�=-3$

$�=-1$

4. $\ln (5)+7�(1-2�)=\ln (18)$

First, distribute the $7�$ on the left side:

$\ln (5)+7�-14��=\ln (18)$

Now, subtract $\ln (5)$ from both sides:

$7�-14��=\ln (18)-\ln (5)$

Using the properties of natural logarithms, we can combine the right side into a single logarithm:

$7�-14��=\ln \left(\frac{18}{5}\right)$

Now, factor out $�$ from the left side:

$�(7-14�)=\ln \left(\frac{18}{5}\right)$

Finally, solve for (x:

$7-14�=\frac{\ln \left(\frac{18}{5}\right)}{�}$

$14�=7-\frac{\ln \left(\frac{18}{5}\right)}{�}$

$�=\frac{7}{14}-\frac{\ln \left(\frac{18}{5}\right)}{14�}$

$�=\frac{1}{2}-\frac{\ln \left(\frac{18}{5}\right)}{14�}$

5. ${\log }_6(�+9)-{\log }_4(�)={\log }_4(49)$

Using the properties of logarithms, we can combine the left side into a single logarithm:

${\log }_6\left(\frac{�+9}{�}\right)={\log }_4(49)$

Now, both sides have the same base, which is 4. According to the one-to-one property of logarithms, if ${\log }_{�}(�)={\log }_{�}(�)$, then $�=�$.

So, $\frac{�+9}{�}=49$

Now, solve for (x:

$�+9=49�$

Subtract $�$ from both sides:

$9=49�-�$

$9=48�$

$�=\frac{9}{48}=\frac{3}{16}$

6. $\ln (2)-\ln (2-2�)=\ln (6)$

First, subtract $\ln (2)$ from both sides:

$-\ln (2-2�)=\ln (6)-\ln (2)$

Now, apply the properties of natural logarithms to the right side:

$-\ln (2-2�)=\ln \left(\frac{6}{2}\right)$

Simplify:

$-\ln (2-2�)=\ln (3)$

Take the exponential