Problem 91 - 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. $$ \ln (x-2)-\ln (x+3)=\ln (x-1)-\ln (x+7) $$

Step-by-Step Solution

Answer

The exact value of the solution to the given logarithmic equation is \(x=3.4\).

Step 1: Application of logarithm properties

Apply the property of logarithms that the difference of logarithms is equivalent to the logarithm of the quotient. The given equation becomes: \(\ln\frac{x-2}{x+3}=\ln\frac{x-1}{x+7}\)

Step 2: Equality of logarithms

The logarithms of two numbers are equal if and only if the numbers themselves are equal. Hence, \(\frac{x-2}{x+3} = \frac{x-1}{x+7}\). This translates into cross multiplication to get rid of the fraction.

Step 3: Cross multiplication

Cross multiply the equation giving, \((x-2)(x+7) = (x+3)(x-1)\) which simplifies to: \(x^2+7x-2x-14=x^2+x-3-x-21\)

Step 4: Simplifying the equation

After simplification the equation will be:\(5x -17=0\) which we will then solve for x.

Step 5: Solving for x

The solution of the equation \(5x-17=0\) is \(x=3.4\)

Step 6: Checking for the domain

While the result of \(x=3.4\) appears to be a solution, we need to verify it against the original logarithmic expressions since logarithms are not defined for negative inputs or zero. Inspecting the original problem, we see that all of the expressions \(x-2\), \(x+3\), \(x-1\), and \(x+7\) must be greater than zero. Checking these for \(x=3.4\), we find this value of \(x\) satisfies all conditions. Therefore, \(x=3.4\) is the solution.

Key Concepts

Logarithm Properties

Understanding the properties of logarithms is crucial when solving logarithmic equations. Logarithms can often seem daunting, but with the right set of rules, they are much more approachable. One of these essential rules states that the difference of logarithms, such as \( \ln(a) - \ln(b) \), is equal to the logarithm of the division of the arguments, expressed as \( \ln\left(\frac{a}{b}\right) \).

Another important property is that if the logarithms of two expressions are equal, such as \( \ln(x) = \ln(y) \), then the expressions themselves must be equal, so \( x = y \). Using these properties correctly is vital in breaking down complex logarithmic equations into solvable algebraic ones. Moreover, recognizing when to apply the properties of logs simplifies the equation and sets the stage for further algebraic manipulations.

Cross Multiplication

Cross multiplication is a technique used to solve equations involving fractions or ratios. It's a straightforward method: if you have two fractions set equal to each other, \( \frac{a}{b} = \frac{c}{d} \), you can multiply the numerator of one fraction by the denominator of the other, and do the same with the remaining numerator and denominator. The equation \( ad = bc \) results from this process.

This method is particularly useful in solving logarithmic equations where you've applied logarithm properties to create ratios. In our exercise, once the logarithms were eliminated, cross multiplication provided a simple algebraic equation that could be solved for the variable \( x \). The beauty of cross multiplication lies in its ability to eliminate the fractions, paving the way for basic algebraic operations like addition, subtraction, multiplication, and division to solve for the unknown.

Logarithmic Domain

The domain of a logarithmic function is an important concept that must be taken into account when solving logarithmic equations. Logarithms are not defined for negative numbers or zero, which means their arguments must be positive. Therefore, the domain of a logarithmic expression like \( \ln(x) \) is \( x > 0 \).

When finding solutions to logarithmic equations, it's essential to check that your solution falls within the domain of the original logarithmic expressions. If your solution results in a negative number or zero inside the log function, it is not valid and must be rejected. In the given exercise, we solved for \( x \), resulting in \( x = 3.4 \). This solution has to be verified against all the original logarithmic expressions to ensure it's within the domain, meaning all the expressions \( x-2 \) , \( x+3 \) , \( x-1 \) , and \( x+7 \) when \( x = 3.4 \) are positive. Proper domain consideration ensures the integrity and correctness of the solution.