Problem 47 - Precalculus Mathematics for Calculus

Question

Solve the logarithmic equation for \(x .\) $$\log x+\log (x-1)=\log (4 x)$$

Step-by-Step Solution

Answer

The solution is \(x = 5\).

Step 1: Combine Logarithmic Terms

Use the logarithmic property that combines sums of logs: \(\log a + \log b = \log(ab)\). This gives us the left side of the equation as \(\log(x(x-1))\). Thus, the equation becomes: \[\log(x(x-1)) = \log(4x)\]

Step 2: Set Arguments Equal

Since both sides of the equation are in log form and their bases are identical, set the arguments equal to each other: \[x(x-1) = 4x\] Simplify to get: \[x^2 - x = 4x\]

Step 3: Rearrange Equation

Rearrange the equation to bring all terms to one side: \[x^2 - x - 4x = 0\] This simplifies to: \[x^2 - 5x = 0\]

Step 4: Factor the Quadratic

Factor out the common term \(x\) from the equation: \[x(x - 5) = 0\] This gives us two potential solutions: \(x = 0\) or \(x = 5\).

Step 5: Validate Solutions

Check the potential solutions in the context of the original equation, considering the domain restrictions of logarithmic functions. Since \(\log(0)\) and \(\log(-1)\) are not defined, \(x = 0\) is not a valid solution. For \(x = 5\), both \(x\) and \(x-1\) are positive, making it a valid solution.

Key Concepts

Logarithmic Properties

Logarithmic properties are crucial tools in solving equations that involve logs. One important property, used in the given exercise, is about combining logarithms. This is known as the product rule of logarithms. It states that the sum of logarithms with the same base can be combined into a single logarithm:

• \(\log a + \log b = \log(ab)\)

This property allows us to simplify complex equations, making it easier to solve for the unknown variable.
For example, in the equation given, \(\log x + \log(x - 1)\) was converted to \(\log(x(x-1))\). This step significantly simplifies the equation, setting the stage to solve it more straightforwardly by comparing arguments.
When utilizing logarithmic properties, ensure that all logarithmic terms have the same base. This consistency is key in applying these properties successfully.

Quadratic Equations

Quadratic equations appear often in algebra, especially following simplifications like we saw in the original exercise. Once the logarithmic expressions were combined and arguments set equal, the resulting equation was a quadratic:

• \(x^2 - 5x = 0\)

Quadratics usually take the form \(ax^2 + bx + c = 0\). They can often be solved by factoring, completing the square, or using the quadratic formula. In this case, factoring was straightforward, as \(x^2 - 5x\) could be factored into \(x(x - 5)\).
The values of \(x\) that satisfy this equation are found by setting each factor to zero:

• \(x = 0\)
• \(x - 5 = 0\)

This gives potential solutions of \(x = 0\) and \(x = 5\). However, it's essential to verify these solutions within the context of the problem to ensure they are valid.

Domain Restrictions in Logarithms

Domain restrictions in logarithmic functions are important to remember, especially when solving equations. For a logarithm \(\log_b(x)\), the argument \(x\) must always be positive. The base of a logarithm also must be positive and not equal to 1, although it's usually implied when not explicitly mentioned.
In our exercise, when checking solutions for \(x\), we found that \(x = 0\) was not valid. This is because \(\log(0)\) is undefined (logarithms of non-positive numbers do not exist in the real number system). Similarly, \(x - 1\) must also be positive because it's part of another logarithmic argument.

• \(x > 0\)
• \(x - 1 > 0\)

These restrictions mean that realistically we can only use solutions that satisfy these conditions, making \(x = 5\) the only valid solution.
Considering these restrictions confirms that not every algebraic solution derived from manipulating an equation is necessarily valid within the original context of the problem.