Solving Exponential & Logarithmic Equations Lesson

Before we jump in and start solving exponential and logarithmic equations, let's look at some of the properties we will be using in this lesson.

Properties for Solving Exponential and Logarithmic Equations

• b, x, and y are real numbers, b > 0, b ≠ 1
• If x = y, then b^x = b^y
• If b^x = b^y, then x = y
• If x = y and x > 0, y > 0, then log_b(x) = log_b(y)
• If x > 0, y > 0 and log_b(x) = log_b(y), then x = y

Solving Exponential Equations with Different Bases

Previously, we learned how to solve exponential equations with the same base. For that scenario, we can set the exponents equal to each other and solve for the unknown. We can't do this with every scenario since it is not always possible to have the same base on each side of an equation. When we try to solve an exponential equation and different bases are involved, we use logarithms to obtain our solution. We will first isolate the exponential expression and then we can take logarithms to the same base on both sides. This will allow us to use our power rule for logarithms to isolate the variable. Let's look at a few examples.
Example 1: Solve each equation. $$6^x=15$$ To solve this equation, we can take the log of each side. $$log(6^x)=log(15)$$ Using our power rule for logarithms, we can bring the exponent (x) down in front of the logarithm on the left side of the equation. $$xlog(6)=log(15)$$ To get x by itself, we divide each side of the equation by log(6). $$x=\frac{log(15)}{log(6)}$$ This is our answer. If the teacher or assignment calls for a decimal approximation, we can divide using a calculator and then round to the appropriate number of decimal places.
One thing to note, the textbook may give you an alternative method for solving this type of equation. Notice that our base of the exponential expression is a 6. Recall the property of logarithms that tells us:
log_b(b^x) = x
Therefore, we can also solve the equation in the following way: $$6^x=15$$ $$log_{6}(6^x)=log_{6}(15)$$ $$x=log_{6}(15)$$ Since using the change of base rule gives us our answer above, we can see that both answers are the same. It is just a matter of personal preference.
Example 2: Solve each equation. $$10 \cdot 19^{4 - x}- 5=77$$ To solve this equation, we want to isolate the exponential expression. This will allow us to use logarithms and bring the variable down out of the exponent. We will begin by adding 5 to each side: $$10 \cdot 19^{4 - x}=82$$ Now we will divide each side by 10: $$19^{4 - x}=\frac{82}{10}$$ We can reduce our fraction on the right: $$19^{4 - x}=\frac{41}{5}$$ Now, we can take the log of each side: $$log(19^{4 - x})=log\left(\frac{41}{5}\right)$$ Bring the exponent (4 - x) out in front: $$(4 - x)log(19)=log\left(\frac{41}{5}\right)$$ Divide each side by log(19): $$4 - x=\frac{log\left(\frac{41}{5}\right)}{log(19)}$$ Subtract 4 away from each side: $$-x=\frac{log\left(\frac{41}{5}\right)}{log(19)}- 4$$ Multiply each side by -1: $$x=-\frac{log\left(\frac{41}{5}\right)}{log(19)}+ 4$$ Example 3: Solve each equation. $$e^{x - 3}- 4=10$$ Let's begin by isolating our exponential expression. We will add 4 to each side of the equation: $$e^{x - 3}=14$$ When e is involved, we want to use ln.
Remember ln is log_e $$ln(e^{x - 3})=ln(14)$$ $$x - 3=ln(14)$$ We will add 3 to each side of the equation: $$x=ln(14) + 3$$

Solving Logarithmic Equations

We will also learn how to solve logarithmic equations.

• Transform the equation so that a single logarithm appears on one side
  • We can use the product rule or quotient rule for logarithms to accomplish this task
• Use one of the following rules to obtain a solution
  • If log_b(x) = log_b(y), then x = y
  • If log_b(x) = k, then x = b^k

Let's look at a few examples.
Example 4: Solve each equation. $$log_{5}(30)=log_{5}(3x + 9)$$ To solve this equation, we use the following property:
If x > 0, y > 0 and log_b(x) = log_b(y), then x = y
Since we have the same base on each log, we can set the arguments equal to each other: $$3x + 9=30$$ Subtract 9 away from each side of the equation: $$3x=21$$ Divide each side by 3: $$x=7$$ Example 5: Solve each equation. $$9 - 3log_{8}(3x - 1)=6$$ For this scenario, we want to isolate the logarithm. Let's begin by subtracting 9 away from each side of the equation. $$-3log_{8}(3x - 1)=-3$$ Divide each side by -3: $$log_{8}(3x - 1)=1$$ To solve this equation, we use the following property:
If log_b(x) = k, then x = b^k
In other words, we write this in exponential form: $$3x - 1=8^1$$ $$3x - 1=8$$ Add 1 to each side of the equation: $$3x=9$$ Divide each side by 3: $$x=3$$

Solving Logarithmic Equations with Extraneous Solutions

When solving logarithmic equations, we may see extraneous solutions. These will occur when a proposed solution results in an undefined logarithm. Let's look at an example.
Example 6: Solve each equation.
$$log_{8}(x - 8) + log_{8}(x - 10)=log_{8}(15)$$ Condense the left side: $$log_{8}[(x - 8)(x - 10)]=log_{8}(15)$$ $$log_{8}(x^{2}- 18x + 80)=log_{8}(15)$$ Set the arguments equal to each other: $$x^{2}- 18x + 80=15$$ $$x^{2}- 18x + 65=0$$ Factor the left side: $$(x - 13)(x - 5)=0$$ $$x=13, 5$$ In this case, only 13 will work as a solution.
Let's begin by substituting 13 in for x: $$log_{8}(13 - 8) + log_{8}(13 - 10)=log_{8}(15)$$ $$log_{8}(5) + log_{8}(3)=log_{8}(15)$$ $$log_{8}(15)=log_{8}(15)$$ When we substitute a 5 in for x, we will see a problem: $$log_{8}(5 - 8) + log_{8}(5 - 10)=log_{8}(15)$$ $$log_{8}(-3) + log_{8}(-5)=log_{8}(15)$$ Although -3 • -5 is 15, our two logarithms on the left-hand side are undefined. This means we must reject our solution of x = 5.

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