Problem 53 - Beginning and Intermediate Algebra

Question

Solve the logarithmic equation algebraically. Then check using a graphing calculator. $$\log (2 x+1)-\log (x-2)=1$$

Step-by-Step Solution

Answer

The solution to the logarithmic equation \(\log (2x + 1) - \log (x - 2) = 1\) can be found by combining the logarithms and solving for \(x\). The combined logarithm is \(\log \frac{2x + 1}{x - 2} = 1\), which gives \(\frac{2x + 1}{x - 2} = 10\). Solving for \(x\) yields \(x = \frac{21}{8}\). A graphing calculator confirms that this value of \(x\) is indeed the correct solution.

Step 1: Combine the logarithms

We'll start by combining the two log functions using the log property: \[\log_ba - \log_bc = \log_b\frac{a}{c}\] Similarly, we get the following equation: \[\log(2x + 1) - \log(x - 2) = \log\frac{2x + 1}{x - 2}\]

Step 2: Apply the given equation

Now, we are given that the difference of the two logs is 1, so: \[\log\frac{2x + 1}{x - 2} = 1\]

Step 3: Remove the logarithm

To remove the logarithm, we can rewrite the equation as: \[\frac{2x + 1}{x - 2} = 10^{1}\] Since the base of the log function is not explicitly given, it implies that the base is 10. Therefore, we raise both sides of the equation to the power of the base (10), resulting in: \[\frac{2x + 1}{x - 2} = 10\]

Step 4: Solve for x

Now we can solve the equation for \(x\): \begin{align*} 2x + 1 &= 10(x - 2) \\ 2x + 1 &= 10x - 20 \\ -8x &= -21 \\ x &= \frac{21}{8} \end{align*}

Step 5: Check the solution using a graphing calculator

To check our solution, we can use a graphing calculator to plot the given equation \[\log (2 x+1)-\log (x-2)=1\] and see if the value \(x = \frac{21}{8}\) is indeed the solution. By plotting the equation, we should see that our solution corresponds to the point where the two functions intersect, and the graph will confirm that \(x = \frac{21}{8}\) is the solution to the given logarithmic equation.

Key Concepts

Properties of Logarithms

Logarithms, often just called logs, have specific properties that help us simplify and solve equations involving them. One such property is the subtraction rule. This is expressed as:

• \(\log_b a - \log_b c = \log_b\left(\frac{a}{c}\right)\)

This means that when you subtract one logarithm from another, you can rewrite the expression as the logarithm of the division of their arguments.
Another important property is the identity that connects logarithms to exponents. For a logarithm to a base \(b\):

• If \(\log_b(x) = y\), then \(b^y = x\)

In the problem above, the base is assumed to be 10 because it is not specified, which is a common practice. Understanding and applying these properties effectively allows you to transform and simplify logarithmic expressions, making it easier to isolate variables and solve equations.

Solving Equations

Solving logarithmic equations involves several steps and often requires explicit manipulation to isolate the variable. For the equation \(\log(2x + 1) - \log(x - 2) = 1\), the goal is to solve for \(x\).
First, combine the log terms using the properties of logarithms:

• \(\log\left(\frac{2x + 1}{x - 2}\right) = 1\)

Next, remove the logarithm by converting it to its equivalent exponential form. This transition is facilitated by the intrinsic relationship between logarithms and exponents:

• \(\frac{2x + 1}{x - 2} = 10^1\)

Finally, solving this equation will determine the value of \(x\). The equation becomes a simpler algebraic equation after removing the logarithms, leading directly to solving using basic algebraic methods.

Algebraic Manipulation

With the logarithmic equation simplified to \(\frac{2x + 1}{x - 2} = 10\), algebraic manipulation can proceed.To solve this, first eliminate the fraction by cross-multiplying:

• \(2x + 1 = 10(x - 2)\)

Here, you distribute and gather like terms to one side:

• \(2x + 1 = 10x - 20\)
• This simplifies to \(-8x = -21\)

Dividing through by \(-8\) gives the solution for \(x\):

• \(x = \frac{21}{8}\)

Checking this solution is prudent, often using a graphing calculator to visually confirm the intersection point of the logarithmic expressions denotes our solution. Once \(x = \frac{21}{8}\) is confirmed graphically, confidence in the solution's accuracy is secured.