Problem 72
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. $$ \log _{4}(x+2)-\log _{4}(x-1)=1 $$
Step-by-Step Solution
Verified Answer
The solution to the logarithmic equation is \(x = 2\).
1Step 1: Apply quotient rule
Begin by applying the quotient rule for logarithms. This rule states that \( \log_b(a) - \log_b(b) = \log_b \left( \frac{a}{b} \right) \). This lets us rewrite the equation \(\log _{4}(x+2)-\log _{4}(x-1) = 1\) as \( \log _{4}\left( \frac{x+2}{x-1} \right) = 1 \).
2Step 2: Remove the Logarithm
To continue solving and to get rid of the logarithm, use the definition of a log that states \( \log_b(a) = c \) is equivalent to \( b^c = a \). Therefore, \(4^1 = \frac{x+2}{x-1}\), which simplifies to \(4 = \frac{x+2}{x-1}\).
3Step 3: Solve for x
Next, solve for x. Multiply both sides of the equation by \(x-1\) to get \(4(x-1) = x+2\). Simplify this to \(4x-4 = x+2\). Solving for x gives \(x = 2\).
4Step 4: Check domain validity
Finally we need to ensure that this solution is in the domain of the original logarithmic expressions. Recall that the argument of a logarithmic expression must always be greater than zero. This means both \(x+2 > 0\) and \(x-1 > 0\) must hold, giving us the domain as \(x > -2\) and \(x > 1\) respectively, which includes \(x = 2\). This validates our solution.
Key Concepts
Quotient Rule for LogarithmsDomain of Logarithmic ExpressionsSolving Equations Step by StepLogarithm Properties
Quotient Rule for Logarithms
At the heart of solving logarithmic equations lies the powerful quotient rule for logarithms. This rule helps us simplify expressions where two logarithms with the same base are subtracted from each other. It states: \( \log_b(a) - \log_b(c) = \log_b \left( \frac{a}{c} \right) \). This means you can combine the two logarithms into a single one by dividing the inner parts.
For example, in the given problem, \( \log_4(x+2) - \log_4(x-1) = 1 \), using the quotient rule lets us combine the terms into \( \log_4 \left( \frac{x+2}{x-1} \right) = 1 \). This step is crucial because it simplifies the equation significantly, reducing the number of steps needed to find the solution.
This is important for tackling various types of logarithmic problems efficiently, especially those involving subtraction of logarithmic terms.
For example, in the given problem, \( \log_4(x+2) - \log_4(x-1) = 1 \), using the quotient rule lets us combine the terms into \( \log_4 \left( \frac{x+2}{x-1} \right) = 1 \). This step is crucial because it simplifies the equation significantly, reducing the number of steps needed to find the solution.
This is important for tackling various types of logarithmic problems efficiently, especially those involving subtraction of logarithmic terms.
Domain of Logarithmic Expressions
The domain of logarithmic expressions is vital to understand because it dictates the acceptable range for the variable. Logarithms only handle positive numbers, which means the expression within the logarithm must always be greater than zero.
In our exercise, the expressions are \(x+2\) and \(x-1\). So, for the logarithm \(\log_4(x+2)\) to exist, \(x+2 > 0\) must be true, yielding \(x > -2\). Similarly, \(x-1 > 0\) translates to \(x > 1\). The stricter of the two, \(x > 1\), determines the domain for our problem.
This checking step ensures we do not end up with a solution that makes any part of the logarithmic expression negative or zero, which would invalidate the solution.
In our exercise, the expressions are \(x+2\) and \(x-1\). So, for the logarithm \(\log_4(x+2)\) to exist, \(x+2 > 0\) must be true, yielding \(x > -2\). Similarly, \(x-1 > 0\) translates to \(x > 1\). The stricter of the two, \(x > 1\), determines the domain for our problem.
This checking step ensures we do not end up with a solution that makes any part of the logarithmic expression negative or zero, which would invalidate the solution.
Solving Equations Step by Step
Breaking down and solving logarithmic equations step by step is the most effective way to tackle seemingly complex problems.
Start by simplifying the equation, often using logarithmic rules. In the given example, after applying the quotient rule, we had \(\log_4 \left( \frac{x+2}{x-1} \right) = 1 \). The next step is to "free" the equation from the logarithm by converting it into an exponential form. This uses the definition that \(\log_b(a) = c\) means \(b^c = a\). Results in \(4^1 = \frac{x+2}{x-1} \).
This transition allows us to deal with the equation algebraically: multiplying through by \(x-1\) yields \(4(x-1) = x+2\), then simplifying to \(4x - 4 = x + 2\). Finally, solve for \(x\) as you would a standard equation, leading to \(x = 2\).
This structured approach ensures we methodically tackle each part of the problem and gradually uncover the solution.
Start by simplifying the equation, often using logarithmic rules. In the given example, after applying the quotient rule, we had \(\log_4 \left( \frac{x+2}{x-1} \right) = 1 \). The next step is to "free" the equation from the logarithm by converting it into an exponential form. This uses the definition that \(\log_b(a) = c\) means \(b^c = a\). Results in \(4^1 = \frac{x+2}{x-1} \).
This transition allows us to deal with the equation algebraically: multiplying through by \(x-1\) yields \(4(x-1) = x+2\), then simplifying to \(4x - 4 = x + 2\). Finally, solve for \(x\) as you would a standard equation, leading to \(x = 2\).
This structured approach ensures we methodically tackle each part of the problem and gradually uncover the solution.
Logarithm Properties
Understanding logarithm properties is essential for solving and transforming logarithmic equations. They provide the tools needed to manipulate and simplify expressions, making the path to the solution clearer.
Some fundamental logarithmic properties include:
In our exercise, the quotient rule simplified two logarithms into a single expression, enabling us to solve the equation more straightforwardly. By mastering these properties, students gain the ability to manipulate and solve logarithmic problems with greater ease.
Some fundamental logarithmic properties include:
- Product Rule: \(\log_b(mn) = \log_b(m) + \log_b(n)\).
- Quotient Rule: \(\log_b \left( \frac{m}{n} \right) = \log_b(m) - \log_b(n)\).
- Power Rule: \(\log_b(m^n) = n\log_b(m)\).
In our exercise, the quotient rule simplified two logarithms into a single expression, enabling us to solve the equation more straightforwardly. By mastering these properties, students gain the ability to manipulate and solve logarithmic problems with greater ease.
Other exercises in this chapter
Problem 71
Use a calculator with an \(\overline{e^{x}}\) key to solve Exercises \(71-76\). Average annual premiums for employer-sponsored family health insurance policies
View solution Problem 72
The figure shows the graph of \(f(x)=\ln x\). Use transformations of this graph to graph each function. Graph and give equations of the asymptotes. Use the grap
View solution Problem 72
In Exercises \(71-78,\) use common logarithms or natural logarithms and a calculator to evaluate to four decimal places. $$ \log _{6} 17 $$
View solution Problem 72
Use a calculator with an \(\overline{e^{x}}\) key to solve Exercises \(71-76\). Average annual premiums for employer-sponsored family health insurance policies
View solution