Problem 45 - Beginning and Intermediate Algebra

Question

Solve the logarithmic equation algebraically. Then check using a graphing calculator. $$\log _{8}(x+1)-\log _{8} x=2$$

Step-by-Step Solution

Answer

The solution to the logarithmic equation \( \log_{8}(x+1) - \log_{8}x = 2 \) is \(x = \frac{1}{63}\). The graphs of \(y_1 = \log_{8}(x+1) - \log_{8}x\) and \(y_2 = 2\) intersect at the point \(\left(\frac{1}{63}, 2\right)\), confirming that our solution is correct.

Step 1: Combine the logarithms

Using the logarithm properties, we can combine the logarithms into a single logarithm by dividing the argument of the second logarithm by the argument of the first logarithm: \( \log _{8}(x+1)-\log _{8} x = \log _{8}\left(\frac{x+1}{x}\right) \) Now, our equation becomes: \( \log _{8}\left(\frac{x+1}{x}\right) = 2 \)

Step 2: Rewrite in exponential form

Now, we need to rewrite the equation in exponential form using the definition of logarithms: \( 8^2 = \frac{x+1}{x} \)

Step 3: Solve for x

Now, we need to solve for \(x\): \(64 = \frac{x+1}{x}\) Multiply by x to get rid of the fraction: \(64x = x+1\) Now subtract x from both sides to isolate the term with x on one side: \(63x = 1\) Finally, divide by 63 to find the value of x: \(x = \frac{1}{63}\)

Step 4: Check the solution using a graphing calculator

To check the solution using a graphing calculator, graph the following functions: \(y_1 = \log _{8}(x+1)-\log _{8} x\) and \(y_2 = 2\) The x-coordinate of the intersection point of the two graphs represents the solution to the equation. You should find that the graph of \(y_1\) intersects the graph of \(y_2\) at the point \(\left(\frac{1}{63}, 2\right)\), confirming that our solution \(x = \frac{1}{63}\) is correct.

Key Concepts

Properties of Logarithms

When solving logarithmic equations, understanding the properties of logarithms is crucial. These properties allow us to manipulate and simplify expressions. Here are some of the most important properties:

• Product Rule: \( \log_b(MN) = \log_bM + \log_bN \) - This rule states that the logarithm of a product is the sum of the logarithms of its factors.
• Quotient Rule: \( \log_b\left(\frac{M}{N}\right) = \log_bM - \log_bN \) - This allows us to express the logarithm of a division as the difference of two logarithms.
• Power Rule: \( \log_b(M^k) = k \cdot \log_bM \) - States that the logarithm of a power is the exponent multiplied by the logarithm of the base.

In the given exercise, we used the Quotient Rule to combine \( \log_8(x+1) - \log_8x \) into a single logarithm: \( \log_8\left(\frac{x+1}{x}\right) \). This simplification step is key in making the equation easier to solve.

Exponential Form

Converting a logarithmic equation to exponential form is a powerful technique in solving equations. The fundamental equivalence between logarithms and exponents is expressed as:

• If \( \log_bA = C \), then \( b^C = A \).

This conversion helps us eliminate the logarithm from the equation, making it a more straightforward algebraic expression to solve. In our example, this meant rewriting \( \log_8\left(\frac{x+1}{x}\right) = 2 \) into exponential form: \( 8^2 = \frac{x+1}{x} \).
This step simplifies the logarithmic equation into a form that can easily be solved using algebraic techniques.

Graphing Calculator

A graphing calculator is a useful tool for visually verifying solutions to equations. It helps confirm our algebraic solution by graphing and identifying where two functions intersect. Here is how you can check the solution:

• First, enter the expression \( \log_8(x+1) - \log_8x \) into the calculator as \((y_1)\).
• Then, plot the constant line \( y_2 = 2 \).

The x-coordinate of the intersection point of these two graphs represents the solution to the original equation. In this case, the point \( \left(\frac{1}{63}, 2\right) \) confirms our calculated solution, \( x = \frac{1}{63} \), providing visual assurance of our results.

Solving Algebraically

Solving algebraically involves manipulating equations to isolate the variable of interest, which usually means applying a series of logical steps and operations. Here's how we proceed after converting the equation to its exponential form:

• First, tackle any fractions by multiplying to clear the denominator. In the equation \( 64 = \frac{x+1}{x} \), multiply every term by \( x \) to eliminate the fraction: \( 64x = x + 1 \).
• Next, bring all terms involving \( x \) to one side: \( 63x = 1 \).
• Finally, divide to solve for \( x \), resulting in \( x = \frac{1}{63} \).

Each of these steps is a foundational algebraic technique. Solving logarithmic equations algebraically requires understanding how to rearrange and unfold the equation step by step, ensuring a solid grasp of algebra principles.