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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. $$\ln (x-5)-\ln (x+4)=\ln (x-1)-\ln (x+2)$$

Short Answer

Expert verified
The exact solution for the given logarithmic equation is \(x = 29/7\), which is approximately equal to \(4.14\) to two decimal places.

Step by step solution

01

Applying the Quotient Rule

Use the quotient property of logarithms to simplify the equation. The quotient property states that \(\ln(a) - \ln(b) = \ln(a/b)\), so the given equation can be rewritten as \(\ln((x-5)/(x+4)) = \ln((x-1)/(x+2))\)
02

Setting the Expressions Equal

Because the natural logarithm is a one-to-one function, the expression \((x-5)/(x+4)\) must be equal to the expression \((x-1)/(x+2)\): \((x-5)/(x+4) = (x-1)/(x+2)\)
03

Solving for x

Cross multiply and simplify to isolate x. After cross multiplying, you'll get the equation \(x^2 -6x + 9 = x^2 - x - 20\), which simplifies to \(x = 29/7\)
04

Checking the Solution

Substitute \(x = 29/7\) back into the original equation to ensure that it fits within the domain of the logarithmic expressions. In this case, all four logarithmic expressions are defined at \(x = 29/7\) so the solution is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithm Properties
Understanding logarithm properties is crucial when solving logarithmic equations. These properties are tools that help us manipulate the expressions into a form that is easier to solve. For example, one of the fundamental properties is the one-to-one nature of logarithms, meaning if \(\log_b(m) = \log_b(n)\), then \(m = n\).

Other essential properties include the Product Rule, which states that \(\log_b(mn) = \log_b(m) + \log_b(n)\), and the Quotient Rule, which simplifies division inside a logarithm into a difference of two logs: \(\log_b(\frac{m}{n}) = \log_b(m) - \log_b(n)\). This rule is particularly helpful and was used in the exercise to combine the logs on each side of the equation into a single logarithmic expression.

Another key property is the Power Rule, which allows us to take an exponent within a logarithm and move it to the front as a coefficient: \(\log_b(m^n) = n\log_b(m)\). Recognizing these properties can significantly expand our ability to handle and solve a variety of logarithmic equations.
Logarithmic Expressions Domain
The domain of a logarithmic expression refers to the range of input values for which the expression is defined. For logarithms, the domain is crucial because the argument of the logarithm—that is, the value inside the logarithm—must always be positive. Since the logarithmic function is undefined for zero and negative inputs, any solution we find for a logarithmic equation must respect this constraint.

For instance, considering the given equation \(\ln (x-5) - \ln (x+4) = \ln (x-1) - \ln (x+2)\), the domain limitations mean that \(x-5 > 0\) and \(x+4 > 0\) for the left side, and \(x-1 > 0\) and \(x+2 > 0\) for the right side. These inequalities tell us that \(x\) must be greater than 5 and -4 (which is satisfied by being greater than 5), and also greater than 1 and -2 (which is satisfied by being greater than 1). As a result, the allowable domain for \(x\) is all values greater than 5. This understanding is critical when verifying if a solution falls within the allowable domain, as we did in Step 4 of the solution process.
Quotient Rule for Logarithms
The Quotient Rule for logarithms is a transformative property that simplifies the process of solving equations involving division within logarithmic functions. As applied in our exercise, the Quotient Rule states that the logarithm of a division is equivalent to the difference of the logarithms: \(\ln(\frac{a}{b}) = \ln(a) - \ln(b)\).

This property effectively reduces the complexity of the given problem by transforming a logarithmic expression containing a division into a difference of two simpler logarithmic expressions. For example, \(\ln (x-5) - \ln (x+4)\) becomes \(\ln(\frac{x-5}{x+4})\).

Using the Quotient Rule is a valuable step in solving logarithmic equations because it often leads to an equation where the inputs of the logarithmic functions are directly comparable, permitting us to then remove the logarithms and solve for the unknown variable, which is what we did in Step 2 of the given problem. This is only possible because of the one-to-one nature of logarithmic functions which signifies that if the logs are equal, their arguments must also be equal. It is by employing this rule that we were able to set up and solve the equation \(\frac{x-5}{x+4} = \frac{x-1}{x+2}\) for \(x\).

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Most popular questions from this chapter

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