Problem 46 - Beginning and Intermediate Algebra

Question

Solve the logarithmic equation algebraically. Then check using a graphing calculator. $$\log x-\log (x+3)=-1$$

Step-by-Step Solution

Answer

The solution to the logarithmic equation \(\log x - \log (x+3) = -1\) is \(x = \frac{1}{3}\). We combined the logarithms, converted to exponential form, and solved for x algebraically. Using a graphing calculator, we confirmed that the graphs of \(y = \log x - \log (x+3)\) and \(y = -1\) intersect at the point \(\left(\frac{1}{3}, -1\right)\), verifying our solution.

Step 1: Combine the logarithms

Using the logarithm property \(\log a - \log b = \log \frac{a}{b}\), we combine the two logarithms: \(\log x - \log (x+3) = \log \frac{x}{x+3}\) So the equation becomes: \[\log \frac{x}{x+3} = -1\]

Step 2: Convert to exponential form

Since the equation is in the form \(\log_{10}(y) = x\), we can convert it to exponential form using the rule \(10^x = y\). Therefore, we have: \[10^{-1} = \frac{x}{x+3}\]

Step 3: Solve for x

Multiply both sides by \(x + 3\) to remove the fraction: \[(x + 3) \cdot 10^{-1} = \frac{x(x + 3)}{x + 3}\] Simplify: \[\frac{1}{10}(x + 3) = x\] Now, distribute the \(\frac{1}{10}\) into the parentheses: \[\frac{1}{10}x + \frac{3}{10} = x\] Subtract \(\frac {1}{10}x\) from both sides to move the variable to the right side: \[\frac{3}{10} = \frac{9}{10}x\] Divide both sides by \(\frac{9}{10}\) to solve for x: \[x = \frac{1}{3}\]

Step 4: Check with a graphing calculator

To confirm our solution, we'll use a graphing calculator to visualize the equation \(\log x - \log (x+3) = -1\). Plot two functions - \(y = \log x - \log (x+3)\) and \(y = -1\). If our solution is correct, the two graphs should intersect at the point \((\frac{1}{3}, -1)\). After graphing the functions, we can see that they do indeed intersect at the point \(\left(\frac{1}{3}, -1\right)\). Therefore, our solution of \(x = \frac{1}{3}\) is correct.

Key Concepts

Logarithm Properties

Understanding logarithm properties is essential when dealing with logarithmic equations. One crucial property used in this exercise is the difference of logarithms:

• \(\log a - \log b = \log \frac{a}{b}\)

This means when you are subtracting one logarithm from another, you can combine them into the logarithm of a quotient. In our exercise, we combined \(\log x\) and \(\log (x+3)\), resulting in the expression \(\log \frac{x}{x+3}\). This step simplifies the overall equation, making it easier to solve.
Remember these properties to make handling logarithms straightforward in your calculations.

Exponential Form

Switching between logarithmic and exponential forms is a handy skill, especially when solving equations. If you have a logarithmic equation like \(\log_{10}(y) = x\), it can be transformed into its exponential counterpart using the rule:

• \(10^x = y\)

In our exercise, after combining logarithms, we obtained \(\log \frac{x}{x+3} = -1\).
By converting this into exponential form, we had \(10^{-1} = \frac{x}{x+3}\). This transformation provides a straightforward path to get rid of the logarithm, simplifying things so we can focus on solving the equation algebraically.
Practice converting between these forms to gain confidence in solving similar equations.

Solving Equations Algebraically

Once the equation is simplified using logarithm properties and converted into exponential form, the next step is solving for the variable algebraically. We started with the expression \(10^{-1} = \frac{x}{x+3}\).

• Multiply both sides by \(x+3\) to eliminate the fraction: \((x+3) \cdot 10^{-1} = x\)
• Distribute \(\frac{1}{10}\) over \(x+3\): \(\frac{1}{10}x + \frac{3}{10} = x\)
• Rearrange to get \(\frac{3}{10} = \frac{9}{10}x\)
• Finally, isolate \(x\) by dividing both sides by \(\frac{9}{10}\): \(x = \frac{1}{3}\)

This structured approach ensures that each step logically follows from the last, leading to an accurate solution.

Graphing Calculators

Graphing calculators are invaluable tools for confirming your algebraic solution. After solving the equation \(x = \frac{1}{3}\), it's reassuring to verify this using technology.
By graphing the functions \(y = \log x - \log (x+3)\) and \(y = -1\) on the calculator, we can visualize where they intersect.

• When the graphs intersect at \(\left(\frac{1}{3}, -1\right)\), it confirms the solution is correct

Graphing calculators simplify complex functions and offer a visual check, enhancing our understanding of the problem. Make sure to familiarize yourself with these devices for successfully solving and checking your work.