Problem 91
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. $$ \ln (x-2)-\ln (x+3)=\ln (x-1)-\ln (x+7) $$
Step-by-Step Solution
Verified Answer
The solution to the logarithmic equation is \(x=11/3\) or approximately \(3.67\).
1Step 1: Applying logarithmic properties
The properties of logarithms can be applied to combine the logs on both sides of the equation into a single log using the quotient rule: \(\ln (x-2) - \ln (x+3) = \ln ((x-2)/(x+3))\) and \(\ln (x-1) - \ln (x+7) = \ln ((x-1)/(x+7))\). The equation becomes: \(\ln ((x-2)/(x+3)) = \ln ((x-1)/(x+7))\).
2Step 2: Equating the arguments
If two logarithms with the same base are equal, then their arguments must also be equal. Therefore: \((x-2)/(x+3) = (x-1)/(x+7)\).
3Step 3: Solving for x
To get rid of the fractional components, cross-multiply: \(x-2)\*(x+7)= (x-1)\*(x+3)\). This simplifies to: \(x^2+5x-14=x^2+2x-3\). Solving for \(x\) involves grouping the like terms and setting the equation to zero, which gives: \(3x=11\). Dividing both sides by 3, we get \(x=11/3\).
4Step 4: Checking domain validity
The original problem asserts that any solution must be valid in the domain of the original logarithmic expression, namely \(x-2>0\), \(x+3>0\), \(x-1>0\), \(x+7>0\). Checking \(x=11/3\) against these conditions, we find that it is valid in all, thus it is a solution to the equation.
Key Concepts
Logarithmic PropertiesDomain of Logarithmic FunctionsExact and Decimal Solutions
Logarithmic Properties
Solving logarithmic equations often requires using key logarithmic properties. These properties help simplify the expressions so you can solve the problem more easily. One such property is the **quotient rule** of logarithms. It states that the logarithm of a quotient is the difference of the logarithms of the numerator and the denominator. For instance:
\[ \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \]
This was used in the exercise to merge the logs on each side of the equation into one. Another useful rule is the ability to equate the arguments of the logarithms when their bases are the same and they equal each other:
- If \( \ln(A) = \ln(B) \) then \( A = B \)
This step reduces the logarithmic equation to an algebraic one, simplifying the process of finding a solution. Knowing these properties and knowing how to apply them is crucial in solving problems involving logarithms.
\[ \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \]
This was used in the exercise to merge the logs on each side of the equation into one. Another useful rule is the ability to equate the arguments of the logarithms when their bases are the same and they equal each other:
- If \( \ln(A) = \ln(B) \) then \( A = B \)
This step reduces the logarithmic equation to an algebraic one, simplifying the process of finding a solution. Knowing these properties and knowing how to apply them is crucial in solving problems involving logarithms.
Domain of Logarithmic Functions
Understanding the domain of logarithmic functions is key to finding valid solutions. The domain refers to all the values that \( x \) can take such that the logarithmic expression is defined. Recall that \( \ln(x) \) is only defined for \( x > 0 \).
When solving a logarithmic equation, you'll need to ensure that any solutions found respect these domain constraints. For example, in the exercise, each part of the expression had constraints:
These inequalities determine that \( x \) must be greater than the largest negative constant involved, which is \( -7 \). The final solution, \( x = \frac{11}{3} \), satisfies all these conditions, thus it is valid.
When solving a logarithmic equation, you'll need to ensure that any solutions found respect these domain constraints. For example, in the exercise, each part of the expression had constraints:
- \(x - 2 > 0\)
- \(x + 3 > 0\)
- \(x - 1 > 0\)
- \(x + 7 > 0\)
These inequalities determine that \( x \) must be greater than the largest negative constant involved, which is \( -7 \). The final solution, \( x = \frac{11}{3} \), satisfies all these conditions, thus it is valid.
Exact and Decimal Solutions
Once you've found an exact solution, sometimes it's useful to express it as a decimal, especially for practical applications. The exact solution in our example was \( x = \frac{11}{3} \).
To convert this to a decimal, you simply divide 11 by 3, resulting in \( 3.67 \) when rounded to two decimal places. It's essential to retain accuracy by using a calculator for such conversions, which ensures precision in your final answer.
Remember: - Exact solutions give you the mathematically precise answer. - Decimal solutions provide an approximation that's often more practical in real-world scenarios.
Both forms are valuable depending on the context of your problem.
To convert this to a decimal, you simply divide 11 by 3, resulting in \( 3.67 \) when rounded to two decimal places. It's essential to retain accuracy by using a calculator for such conversions, which ensures precision in your final answer.
Remember: - Exact solutions give you the mathematically precise answer. - Decimal solutions provide an approximation that's often more practical in real-world scenarios.
Both forms are valuable depending on the context of your problem.
Other exercises in this chapter
Problem 90
determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. The functions \(f(x)=\le
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Determine whether each equation is true or false. Where possible, show work to support your conclusion. If the statement is false, make the necessary change(s)
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In Exercises 81–100, evaluate or simplify each expression without using a calculator. $$ \ln \frac{1}{e^{6}} $$
View solution Problem 92
In Exercises 81–100, evaluate or simplify each expression without using a calculator. $$ \ln \frac{1}{e^{7}} $$
View solution