Exponential and Logarithmic Equations and Inequalities - AP PreCalculus Study Guide

The Core Idea: Exponential and Logarithmic Equations and Inequalities

This topic focuses on the algebraic techniques required to solve for a variable when it appears within an exponent or as the argument of a logarithm. The central challenge is to "undo" the exponential or logarithmic function to isolate the variable. This is accomplished by applying the inverse operation: we use logarithms to solve exponential equations and exponentiation to solve logarithmic equations.

The process involves strategic manipulation using the properties of exponents and logarithms. For exponential equations, this may involve creating a common base, isolating the exponential term and taking a logarithm of both sides, or taking logarithms of both sides directly. For logarithmic equations, strategies include condensing multiple logarithmic terms into a single logarithm, isolating the logarithm and then exponentiating, and ensuring that any potential solution is valid within the domain of the original logarithmic expressions. These same principles are extended to inequalities, with the added consideration of how inequalities behave when these operations are applied.

Key Rules and Properties

The methods for solving exponential and logarithmic equations are based on their inverse relationship and the one-to-one property of their corresponding functions.

Solving Exponential Equations

1. Isolating the Exponential and Taking the Logarithm (EK 2.13.A.1)

   • For an equation with a single exponential expression, such as $A\cdot b^{f(x)}+C=D$, the first step is to algebraically isolate the exponential term $b^{f(x)}$.

   • Once isolated, $b^{f(x)}=M$, take a logarithm of both sides (natural log, $\ln$, or common log, $\log$, are typical choices).

   • ${\log }_c(b^{f(x)})={\log }_c(M)$

   • Using the power property of logarithms, this becomes $f(x)\cdot {\log }_c(b)={\log }_c(M)$, which can then be solved for $x$.

2. Same Base Property (EK 2.13.A.2)

   • For an equation with two exponential expressions that have the same base, $b^{f(x)}=b^{g(x)}$.

   • Since exponential functions are one-to-one, if the outputs are equal and the bases are the same, the exponents must be equal.

   • Set the exponents equal to each other and solve: $f(x)=g(x)$.

3. Different Bases Property (EK 2.13.A.3)

   • For an equation with exponential expressions on both sides but with different bases, such as $a^{f(x)}=b^{g(x)}$.

   • Take a logarithm of both sides of the equation.

   • $\ln (a^{f(x)})=\ln (b^{g(x)})$

   • Apply the power property of logarithms to bring the exponents down as coefficients.

   • $f(x)\cdot \ln (a)=g(x)\cdot \ln (b)$

   • Solve the resulting equation for $x$.

Solving Logarithmic Equations

1. Isolating the Logarithm and Exponentiating (EK 2.13.B.1)

   • For an equation with a single logarithmic expression, such as $A\cdot {\log }_b(f(x))+C=D$, first isolate the logarithmic term ${\log }_b(f(x))$.

   • Once isolated, ${\log }_b(f(x))=M$, rewrite the equation in its equivalent exponential form.

   • $b^M=f(x)$

   • Solve the resulting equation for $x$.

2. Same Base Logarithm Property (EK 2.13.B.2)

   • For an equation with a single logarithmic expression of the same base on each side, ${\log }_b(f(x))={\log }_b(g(x))$.

   • Since logarithmic functions are one-to-one, if the outputs are equal and the bases are the same, the arguments must be equal.

   • Set the arguments equal to each other and solve: $f(x)=g(x)$.

3. Condensing Logarithmic Expressions (EK 2.13.B.3)

   • If an equation contains multiple logarithmic terms, use the properties of logarithms (product, quotient, and power rules) to condense them into a single logarithmic expression.

   • For example, ${\log }_b(f(x))+{\log }_b(g(x))=C$ becomes ${\log }_b(f(x)\cdot g(x))=C$.

   • Once condensed into a single logarithm, solve using the exponentiating method (Rule 1 above).

Understanding Domain Restrictions

A critical nuance in this topic, specifically for logarithmic equations and inequalities, is the concept of the domain of a logarithmic function (EK 2.13.B.4, 2.13.C.2). The argument of any logarithm must be strictly positive. For an expression ${\log }_b(M)$, the condition $M>0$ must always be true.

When solving a logarithmic equation, the algebraic steps (such as squaring a variable or exponentiating) can introduce solutions that are not valid in the context of the original equation. These are known as extraneous solutions.

Therefore, it is an essential and required step to check every potential solution by substituting it back into the original logarithmic equation or inequality. Any solution that results in the argument of a logarithm being zero or negative must be discarded.

For example, if solving $\log (x)+\log (x-3)=1$, you must first establish the domain. We need $x>0$ AND $x-3>0$, which simplifies to $x>3$. If your algebraic process yields solutions $x=5$ and $x=-2$, you must check them against this domain.

• $x=5$ is a valid solution because $5>3$.

• $x=-2$ is an extraneous solution because it is not greater than 3. It would result in $\log (-2)$ and $\log (-5)$, which are undefined in the real number system.

Core Concepts & Rules

• Solving Exponential Equations: Your primary goal is to get the variable out of the exponent.

  • If you can write both sides of the equation with the same base, do so and then set the exponents equal.

  • If bases are different, isolate the exponential term and then take the logarithm of both sides to bring the exponent down.

• Solving Logarithmic Equations: Your primary goal is to get the variable out of the logarithm's argument.

  • If the equation has multiple logarithms on one side, use logarithm properties to condense them into a single logarithm.

  • Once you have a single logarithm isolated (${\log }_b(A)=C$), convert it to exponential form ($b^C=A$) to solve.

  • If you have a single logarithm on each side with the same base (${\log }_b(A)={\log }_b(B)$), set the arguments equal ($A=B$).

• The Mandatory Check: For any logarithmic equation or inequality, you must check your final solutions. Substitute them into the original equation to ensure that all arguments of the logarithms are positive. Any solution that fails this check is extraneous and must be rejected.

• Solving Inequalities: Use the same algebraic techniques as for equations. Remember that for an increasing function $f(x)$ (like $b^x$ or ${\log }_b(x)$ for $b>1$), the inequality direction is preserved when applying the function or its inverse. For example, if $A<B$, then ${\log }_b(A)<{\log }_b(B)$. Always consider the domain restrictions for logarithmic inequalities from the very beginning.

Step-by-Step Example 1: Solving a Logarithmic Equation with Condensing

Problem: Find all real solutions to the equation $\ln (x+5)+\ln (x)=\ln (14)$.

Step 1: Determine the domain of the equation.

For the logarithmic expressions to be defined, their arguments must be positive.

• $x+5>0⟹x>-5$

• $x>0$

Both conditions must be true, so we take the more restrictive condition. The domain for this equation is $x>0$. Any solution we find must satisfy this.

Step 2: Condense the logarithmic expressions.

The left side of the equation is a sum of two logarithms with the same base ($e$). Use the product property of logarithms ($\ln (A)+\ln (B)=\ln (AB)$) to combine them.

$$\ln ((x+5)\cdot x)=\ln (14)$$

Step 3: Apply the one-to-one property of logarithms.

Now the equation has the form $\ln (M)=\ln (N)$. Since the bases are the same, we can set the arguments equal to each other.

$$(x+5)x=14$$

Step 4: Solve the resulting equation.

The equation is a quadratic. Distribute and set it to zero to solve.

$$x^2+5x=14$$

$$x^2+5x-14=0$$

Factor the quadratic expression.

$$(x+7)(x-2)=0$$

This gives two potential solutions: $x=-7$ and $x=2$.

Step 5: Check for extraneous solutions.

We must check both potential solutions against the domain we established in Step 1 ($x>0$).

• Check $x=2$: Is $2>0$? Yes. This is a valid solution.

• Check $x=-7$: Is $-7>0$? No. This is an extraneous solution. If we were to plug it into the original equation, we would get $\ln (-2)$ and $\ln (-7)$, which are undefined.

Final Answer: The only solution is $x=2$.

Step-by-Step Example 2: Solving an Exponential Equation with Different Bases

Problem: Solve the equation $3^{2x-1}=7^x$ for $x$. Provide an exact answer.

Step 1: Take the logarithm of both sides.

Since the bases (3 and 7) are different and cannot be easily rewritten as powers of the same number, we cannot equate the exponents directly. The strategy is to take a logarithm of both sides. The natural logarithm ($\ln$) is a convenient choice.

$$\ln (3^{2x-1})=\ln (7^x)$$

Step 2: Apply the power property of logarithms.

Use the property $\ln (A^B)=B\cdot \ln (A)$ to bring the exponents down in front of the logarithmic terms as coefficients.

$$(2x-1)\ln (3)=x\ln (7)$$

Step 3: Distribute and collect terms with $x$.

Our goal is to isolate $x$. First, distribute $\ln (3)$ on the left side.

$$2x\ln (3)-\ln (3)=x\ln (7)$$

Now, move all terms containing $x$ to one side of the equation and all other terms to the other side.

$$2x\ln (3)-x\ln (7)=\ln (3)$$

Step 4: Factor out $x$ and solve.

Factor $x$ out from the terms on the left side.

$$x(2\ln (3)-\ln (7))=\ln (3)$$

Finally, divide by the coefficient of $x$ to solve for $x$.

$$x=\frac{\ln (3)}{2\ln (3)-\ln (7)}$$

This is the exact solution. Using logarithm properties, the denominator can also be written as $\ln (3^2)-\ln (7)=\ln (9)-\ln (7)=\ln (9/7)$, so an equivalent form of the answer is $x=\frac{\ln (3)}{\ln (9/7)}$.

Final Answer:$x=\frac{\ln (3)}{2\ln (3)-\ln (7)}$

Using Your Calculator

While the AP exam requires you to show analytical solving methods, a graphing calculator is a powerful tool for verifying solutions or solving equations that are difficult or impossible to solve by hand. The primary method is to find the intersection of two functions.

Problem: Find the approximate solution(s) to $2^x=x^2+2$.

Step-by-Step Calculator Guide (TI-84 Style):

1. Enter the Functions:

   • Press the Y= button.

   • In $Y_1$, enter the left side of the equation: $2^X$.

   • In $Y_2$, enter the right side of the equation: $X^2+2$.

2. Graph the Functions:

   • Press GRAPH. You may need to adjust the viewing window to see the points of intersection. Press WINDOW to change $Xmin$, $Xmax$, $Ymin$, and $Ymax$. A standard window (ZOOM -> 6:ZStandard) might not show all solutions. For this problem, setting $Xmax$ to 5 is helpful.

3. Find the Intersection Point(s):

   • Press 2nd then TRACE to open the CALC (calculate) menu.

   • Select 5: intersect.

   • The calculator will ask for the "First curve?". The cursor will be on $Y_1$. Press ENTER.

   • It will then ask for the "Second curve?". The cursor will jump to $Y_2$. Press ENTER.

   • It will ask for a "Guess?". Use the arrow keys to move the cursor close to one of the visible intersection points and press ENTER.

   • The calculator will display the coordinates of the intersection. The $x$-value is the solution to the equation.

4. Repeat for Other Solutions:

   • This particular equation has two points of intersection. Repeat Step 3, but for the "Guess?", move the cursor close to the second intersection point to find the other solution.

For this example, the calculator would find solutions at approximately $x\approx -1.576$ and $x\approx 2.595$.

AP Exam Quick Hit

Common Question Types

• Solving Logarithmic Equations Requiring Condensing and Checking for Extraneous Solutions: You will be given an equation with multiple log terms that must be combined before solving. The key skill tested is identifying and discarding extraneous solutions.

  • Example: "Find the solution(s) to ${\log }_4(x)+{\log }_4(x-6)=2$."

• Solving Exponential Equations with Unlike Bases: These questions require taking the logarithm of both sides and using log properties to isolate the variable, resulting in an exact answer involving logarithms.

  • Example: "Find the exact value of $x$ for which $5^{x+1}=8^{2x}$."

• Solving Logarithmic Inequalities: These problems test both the algebraic steps and the understanding of domain restrictions. The final answer is often an interval.

  • Example: "Find the set of all real numbers $x$ that satisfy the inequality ${\log }_2(3x-1)<3$."

Common Mistakes

• Forgetting to Check for Extraneous Solutions: This is the most frequent error on logarithmic equation problems. Students perform the algebra correctly but fail to plug the potential solutions back into the original equation to ensure all logarithm arguments are positive.

• Incorrectly Applying Logarithm Properties: A common algebraic mistake is misremembering the log rules. For example, incorrectly simplifying $\log (x)-\log (y)$ to $\frac{\log (x)}{\log (y)}$ instead of the correct $\log (\frac{x}{y})$. Another is treating $\log (x+y)$ as $\log (x)+\log (y)$, which is incorrect.

• Isolating Terms Incorrectly: When solving an equation like $5\cdot e^{2x}=20$, a student might incorrectly take the natural log of both sides immediately, writing $\ln (5)+2x=\ln (20)$. The correct first step is to isolate the exponential term by dividing by 5: $e^{2x}=4$.

• Ignoring the Implicit Domain in Inequalities: When solving ${\log }_2(x-5)<3$, a student might correctly solve $x-5<2^3$, which gives $x<13$. However, they often forget the initial domain constraint that $x-5>0$, meaning $x>5$. The correct final answer is the intersection of these two conditions: $5<x<13$.

• Errors in Converting Forms: A simple but common mistake is incorrectly converting between logarithmic and exponential form. Given ${\log }_b(A)=C$, students might write $C^b=A$ or $b^A=C$ instead of the correct $b^C=A$.