Problem 61 - Precalculus with Limits

Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. \(\log _{4} x-\log _{4}(x-1)=\frac{1}{2}\)

Step-by-Step Solution

Answer

The exact solution is 2 and the approximate solution to three decimal places is 2.000.

Step 1: Apply logarithm properties

First, apply the quotient rule of logarithms which states that \(\log _{b} a-\log _{b} c=\log _{b}\left(\frac{a}{c}\right)\) to simplify the equation and bring it to one term.\So, \(\log _{4} x-\log _{4}(x-1)\) becomes \(\log _{4}\left(\frac{x}{x-1}\right)\). Now, the equation is \(\log _{4}\left(\frac{x}{x-1}\right)=\frac{1}{2}.\)

Step 2: Convert from logarithmic to exponential form

Now, convert the equation from logarithmic form to exponential form.Recall that for any b > 0, a > 0, and a ≠ 1, the equation \(\log_{b}{a} = c\) can be written as \(b^{c} = a\).Applying this principle to our equation gives \(4^{0.5}=\frac{x}{x-1}\). Simplifying the left side gives the equation \(2=\frac{x}{x-1}\).

Step 3: Solving the resulting equation

Solve the equation \(2=\frac{x}{x-1}\) for x. First, multiply each side of the equation by \(x - 1\) to get rid of the fraction. This gives \(2(x - 1) = x\).Next, distribute the 2 to get \(2x - 2 = x\). Deduct 'x' from both sides to get \(2x - x = 2\). This simplifies to \(x = 2\). This is the exact solution.

Step 4: Checking the solution

The solution must make the original equation true and be in the domain of the original logarithms. We need to ensure \(x>0\) since we cannot have a log of a nonpositive number and \(x \neq 1\) because that would make the denominator zero in the log term \(\log_{4}\left(\frac{x}{x-1}\right)\). Substituting x = 2 back into the original logarithmic equation and checking gives \( \frac{1}{2} = \frac{1}{2} \), hence x = 2 is the correct solution.

Step 5: Approximate solution

The exact value solution for 'x' is 2. However, the exercise also requests the answer to be approximated to three decimal places. As 2 already meets this criteria, the approximate solution remains 2.000

Key Concepts

Logarithm Properties

Understanding logarithm properties is essential for solving logarithmic equations. One fundamental property frequently used is the quotient rule. It allows you to combine two logarithms with the same base into a single term by expressing them as the logarithm of a quotient. For example, according to the quotient rule, if you have an expression like \(\log_b{a} - \log_b{c}\), it simplifies to \(\log_b{\left(\frac{a}{c}\right)}\).

In the given exercise, the application of this rule turned two separate logarithm terms \(\log_4{x}\) and \(\log_4{(x-1)}\) into \(\log_4{\left(\frac{x}{x-1}\right)}\). By understanding and applying logarithm properties like this, equations become more manageable and solvable using algebraic manipulation.

Exponential Form Conversion

A logarithmic equation can be converted into exponential form to simplify the solving process. It's the essence of what logarithms represent: converting multiplicative relationships into additive ones. The conversion relies on the definition of a logarithm: for \(\log_b{a} = c\), it is equivalent to \(b^c = a\).

In this exercise, you notice an application of this conversion when the logarithmic equation \(\log_4{\left(\frac{x}{x-1}\right)} = \frac{1}{2}\) is rephrased as an exponential equation: \(4^{\frac{1}{2}} = \frac{x}{x-1}\), which simplifies to \(2 = \frac{x}{x-1}\) due to \(4^{\frac{1}{2}}\) being the square root of 4, equal to 2. This transformation from logarithmic to exponential form is a pivotal step that aids in solving for the variable.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying equations to solve for an unknown variable. It involves operations like expanding, factoring, distributing, and combining like terms, as well as applying inverse operations to isolate the variable you’re solving for.

In our example, after converting to exponential form and arriving at \(2 = \frac{x}{x-1}\), the next step is to perform algebraic manipulation to solve for 'x'. Initially, the equation involves a fraction; multiplying each side by \(x - 1\) clears the fraction, leading to \(2(x - 1) = x\). After distributing the 2 and then moving the 'x' terms to the same side, you get a simple equation that can be solved for 'x': \(x = 2\). It's important to always check your solution in the original equation to ensure it's valid, considering the domain restrictions of logarithmic functions.