Problem 84 - College Algebra

Question

Solve each logarithmic equation in Exercises \(49-92 .\) Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ \log (x+7)-\log 3=\log (7 x+1) $$

Step-by-Step Solution

Answer

The solution to the equation is \(x = \frac{1}{2} = 0.50\)

Step 1: Simplify the equation using logarithmic rule

Use the rule \(\log_a b - \log_a c = \log_a \frac{b}{c}\) to simplify the equation. The given equation is \(\log (x+7)-\log 3=\log (7 x+1)\). Applying the rule, the equation becomes \(\log \frac{(x + 7)}{3} = \log (7x + 1)\).

Step 2: Extract values of \(x\)

Since both sides of the equation are equal and have the same base, this implies \(\frac{(x + 7)}{3} = 7x + 1\). Solving for \(x\), yields \(x = \frac{1}{2}\).

Step 3: Validate the solution

Substitute \(x = \frac{1}{2}\) into the original equation to ensure it satisfies all logarithmic terms. Substituting \(x\) into \(x + 7 , 7x + 1\) and checking if all numbers are positive (because log of negative number is undefined).

Step 4: Calculate decimal approximation

If necessary, calculate the decimal approximation of the solution to two decimal places. In this instance, \(x = \frac{1}{2} = 0.50\) after rounding to two decimal places.

Key Concepts

Logarithmic Properties

Logarithms have special properties that make it possible to manipulate and solve equations featuring them. One such property allows you to subtract two logs with the same base, combining them into a single logarithm. In the equation \(\log(x+7)-\log 3=\log(7x+1)\), we use the property \(\log_a b - \log_a c = \log_a \frac{b}{c}\) to simplify the expression. Why is this useful? It transforms the equation into a form that can later be solved for \(x\), by allowing us to equate the arguments of the logarithms (the expressions inside the logarithms).

Remember, properties like this are designed for logarithms of the same base, which is often 10 (common log) or \(e\) (natural log) by default if no base is shown. This exercise particularly emphasizes the importance of understanding and applying these properties to solve logarithmic equations.

Extracting Values from Logarithms

Once logarithmic equations are simplified using properties of logarithms, the next step is to extract the values from them. If an equation is reduced to the form \(\log_a X = \log_a Y\), where \(X\) and \(Y\) are expressions involving \(x\), you can deduce that \(X = Y\) because the logarithm function is one-to-one.

In our problem, after applying the logarithmic subtraction property, we find that \(\log \frac{(x + 7)}{3} = \log (7x + 1)\), which further implies that \(\frac{(x + 7)}{3} = 7x + 1\). Solving this last equation for \(x\) enables us to extract the value of \(x\). However, it is crucial to later verify that the value of \(x\) does not violate the domain restrictions of the log function.

Logarithm Domain Restrictions

Logarithms are only defined for positive real numbers. This is why, when you substitute back your solution into the original equation, it is crucial to check that the arguments of all logarithmic expressions are positive.

In our example, we confirm that when \(x = \frac{1}{2}\), the expressions \(x+7\) and \(7x+1\) yield positive results, thus respecting the domain of the logarithm function. If any logarithmic term would have been negative or zero, our proposed solution \(x\) would have been invalid, regardless of it seemingly solving the algebraic part of the equation.

Decimal Approximation of Solutions

Exact solutions to logarithmic equations are often irrational or complex numbers, but in practical applications, a decimal approximation can be more useful. After solving the equation and verifying that the solution is within the domain constraints, we can use a calculator to find a decimal approximation.

For \(x = \frac{1}{2}\), unless it can be represented as a finite decimal, we typically round it to a specified number of decimal places. Here, the approximation to two decimal places is \(0.50\), which is sufficiently precise for many everyday applications and easier to understand at a glance. The discipline of carrying out these approximations also helps students to translate exact mathematical forms into numbers that carry real-world significance.