Problem 54 - Precalculus with Limits

Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. \(4 \log (x-6)=11\)

Step-by-Step Solution

Answer

The solution of the given logarithmic equation \(x ≈ 168.589\)

Step 1: Isolate the logarithmic term

Isolate the logarithmic term on one side of the equation. Here, we divide both sides of \(4\log(x-6)=11\) by 4 to isolate \(\log(x-6)\), yielding \(\log(x-6) = \frac{11}{4}\)

Step 2: Convert form

Convert the equation from logarithmic form to exponential form. Based on the logarithmic conversion rule \(b=\log_a (x) \rightarrow a^b=x\), we can write the equation \(x-6 = 10^{\frac{11}{4}}\) because the base in the case of common logarithm (log) is 10.

Step 3: Solve for x

Re-arrange the equation obtained in step 2 to solve for x. We do this by adding 6 on both sides, yielding \(x = 6 + 10^{\frac{11}{4}}\).

Step 4: Find numerical answer and round to three decimal places

Using a scientific calculator, compute the numerical value of \(x = 6 + 10^{\frac{11}{4}}\) and round it to three decimal places.

Key Concepts

Solving Logarithmic Equations

Understanding how to solve logarithmic equations is a crucial skill in algebra, as it allows us to find the value of an unknown variable within a logarithm. The first step is to isolate the logarithmic term. In other words, we move all the terms without logarithms to the other side of the equation. For instance, in our example, we start with the equation 4 \(log(x-6)=11\). By dividing both sides by 4, we get \(log(x-6) = \frac{11}{4}\), which simplifies the equation considerably.

Once the logarithmic term is isolated, the next step involves converting the logarithm to its exponential form, which will be discussed in more detail in the next section. After conversion, solving for the variable typically requires basic algebraic operations. For example, if we have \(x - 6 = 10^{\frac{11}{4}}\), we solve for x by adding 6 to both sides. Finally, the calculation is executed, and the result is rounded to the required number of decimal places.

It's important for students to remember to check their answers by substituting the solution back into the original equation. This step confirms whether the solution is correct, particularly because some transformations may introduce extraneous solutions that do not satisfy the original equation.

Converting Logarithmic to Exponential Form

The process of converting logarithmic equations to exponential form is an essential step in solving them. The conversion is based on the fundamental definition of a logarithm: if \(b = \log_a(x)\), then the equivalent exponential form is \(a^b = x\). In this case, a is the base of the logarithm, b is the logarithm itself, and x is the argument of the logarithm.

In the example \(log(x-6) = \frac{11}{4}\), we are dealing with a common logarithm, which means the base a is 10. Subsequently, the exponential form becomes \(10^{\frac{11}{4}} = x-6\). Mastery of this conversion allows for a seamless transition from the realm of logarithms, often seen as complex and abstract, to the more intuitive realm of exponential functions. As such, students should practice this conversion to gain fluency and confidence.

Logarithm Properties

Logarithms have several properties that make them easier to work with, especially when solving equations. Understanding these properties can simplify complex logarithmic expressions and aid in finding solutions efficiently.

Some of the key properties include:

• The Product Rule: \(\log_a(xy) = \log_a(x) + \log_a(y)\), allowing us to separate a logarithm of a product into a sum of logarithms.
• The Quotient Rule: \(\log_a\left(\frac{x}{y}\right) = \log_a(x) - \log_a(y)\), which does the opposite by breaking down a logarithm of a quotient into a difference.
• The Power Rule: \(\log_a(x^p) = p\cdot\log_a(x)\), through which a logarithm with an exponent can be rewritten as a multiple of a logarithm.
• Change of Base Formula: Used to convert logarithms from one base to another, it is expressed as \(\log_b(x) = \frac{\log_a(x)}{\log_a(b)}\).

These properties often come into play when transforming and solving logarithmic equations. For instance, in the exercise 4 \log(x-6)=11, we actually used a reverse application of the Power Rule to go from \(\log(x-6)^4\) to 4 \log(x-6)\. Appreciating these properties helps students to simplify and solve more intricate logarithmic equations effectively.