Problem 83 - Algebra and Trigonometry

Question

Solve each logarithmic equation. 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. $$ \log (x+4)-\log 2=\log (5 x+1) $$

Step-by-Step Solution

Answer

The solution to the logarithmic equation is \(x = 3.6\)

Step 1: Combine logarithms

Combine \(\log (x+4)\) and \(\log 2\) on the left side of the equation using the property \(\log a - \log b = \log (a/b)\). \(\log (x+4) - \log 2 = \log \((x+4)/2\) = \log ((5x + 1)/5)\). This simplifies our equation to \(\log ((x+4)/2) = \log ((5x + 1)/5)\).

Step 2: Remove logarithms

Remember that if \(\log a = \log b\), then it must be true that \(a = b\). We can apply this to our current equation and get: \((x+4)/2 = (5x + 1)/5\).

Step 3: Simplify the equation

First, get rid of the fractions by multiplying each side by the least common multiple of 2 and 5, which is 10. This results in \(5(x + 4) = 2(5x + 1)\). Expand it to get \(5x + 20 = 10x + 2 \).

Step 4: Solve for x

Rearrange the equation to isolate \(x\) on one side: \(5x - 10x = 2 - 20\), or \(-5x = -18\). Solve for \(x\) by dividing both sides by -5, which gives: \(x = 18/5 = 3.6\).

Step 5: Validate the solution

Insert the calculated value of \(x\) into the original logarithmic expressions to ensure they are defined: \(\log (x+4) = \log (3.6 + 4) = \log 7.6\) and \(\log (5x + 1) = \log (5*3.6 + 1) = \log (18 + 1) = \log 19\) - both logarithmic expressions are defined for \(x = 3.6\), so the solution is valid.

Key Concepts

Logarithm Properties

Understanding the properties of logarithms is vital when solving logarithmic equations. One critical property is the subtraction rule, which allows us to combine logarithms with the same base by converting their difference into the logarithm of the division of the arguments, expressed as \( \log a - \log b = \log (\frac{a}{b}) \). Another essential property is the equality of logs, which states that if \( \log a = \log b \), then \( a = b \). These properties enable us to transform logarithmic equations into more solvable algebraic forms.

Additionally, we use properties such as the change of base formula and the product rule \( \log (ab) = \log a + \log b \) when confronting more complex equations. It's also important to remember that logarithms are only defined for positive numbers, as you cannot take the logarithm of zero or a negative number.

Equation Simplification

The process of simplifying an equation is one of systematically performing mathematical operations to achieve an easier-to-solve form of the original equation. When faced with logarithmic equations, simplification often involves removing the logarithms to create a basic algebraic equation.

As seen in the provided exercise, simplification includes combining logarithms into a single log, eliminating fractions by finding a common denominator, and executing basic algebraic operations like expansion and factoring. Simplifying equations step by step reveals the algebraic structure and makes the solution more approachable.

Domain of Logarithmic Functions

The domain of a logarithmic function includes all the values that 'x' can take on without causing the function to become undefined.

For the function \( \log(x) \), the domain is all positive real numbers \( (x > 0) \). It is because the logarithm of a negative number or zero is undefined in the real number system. Therefore, when solving logarithmic equations, it's crucial to check that our answers fall within the domain of the given functions, otherwise, they must be rejected as invalid solutions, as they do not meet the necessary conditions of the original logarithmic expressions.

Decimal Approximation

Obtaining a decimal approximation involves converting an exact solution, often expressed in fractions or irrational numbers, into a decimal. This is usually done with the aid of a calculator.

However, it's not just about pushing buttons; understanding when and why a decimal approximation is necessary is part of the mathematical process. We often seek decimal approximations for more intuitive understanding, for measurement purposes, or to meet a required degree of accuracy. In our exercise, the exact answer was \( \frac{18}{5} \), and its decimal approximation to two decimal places is 3.60. This rounded value is often easier to comprehend and is more practical from an applied perspective.