Solving Logarithmic Equations Lesson

Before we jump in and start solving 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 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 x > 0, y > 0 and 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 1: 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 2: 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: $$x=3$$

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