Problem 94
Question
Solve: \(\log _{4}(x+3)-\log _{4}(x-1)=2\).
Step-by-Step Solution
Verified Answer
The solution is \(x = \frac{19}{15}\).
1Step 1: Combine Logarithms
Use the logarithm subtraction rule: \(\text{log}_{a}(b) - \text{log}_{a}(c) = \text{log}_{a} \frac{b}{c} \) to combine the logarithms on the left side of the equation. \(\text{log}_{4}(x+3) - \text{log}_{4}(x-1) = \text{log}_{4} \frac{x+3}{x-1}\). Therefore, the equation becomes \(\text{log}_{4} \frac{x+3}{x-1} = 2\).
2Step 2: Exponentiate Both Sides
Exponentiate both sides of the equation to eliminate the logarithm using the fact that if \(\text{log}_{a}(b) = c\), then \(\text{b} = a^{c}\). This gives \(\frac{x+3}{x-1} = 4^2\).
3Step 3: Simplify
Simplify \(4^2\) to obtain \(16\). Hence, the equation now is \(\frac{x+3}{x-1} = 16\).
4Step 4: Solve for x
Multiply both sides by \(x-1\) to clear the fraction: \((x+3) = 16(x-1)\). Then distribute and simplify: \(x + 3 = 16x - 16\). Solving for \(x\), we get: \(3 + 16 = 16x - x\) \(19 = 15x\) \(x = \frac{19}{15}\).
Key Concepts
logarithm propertiessolving equationsexponentiation
logarithm properties
Logarithms are powerful tools in mathematics that allow us to solve equations involving exponentiation. One of the most useful properties of logarithms is the ability to combine or split logarithms using their rules.
The subtraction rule is particularly handy. It states: \(\text{log}_{a}(b) - \text{log}_{a}(c) = \text{log}_{a} \frac{b}{c}\).
This rule helps us simplify equations by condensing multiple logarithms into a single expression. In our exercise, we used this subtractive property to combine \(\text{log}_{4}(x+3)\) and \(\text{log}_{4}(x-1)\) into \(\text{log}_{4} \frac{x+3}{x-1}\).
Understanding these properties is crucial as they simplify complex problems and make them more manageable. Thus, mastering these properties will greatly enhance your algebra skills.
The subtraction rule is particularly handy. It states: \(\text{log}_{a}(b) - \text{log}_{a}(c) = \text{log}_{a} \frac{b}{c}\).
This rule helps us simplify equations by condensing multiple logarithms into a single expression. In our exercise, we used this subtractive property to combine \(\text{log}_{4}(x+3)\) and \(\text{log}_{4}(x-1)\) into \(\text{log}_{4} \frac{x+3}{x-1}\).
Understanding these properties is crucial as they simplify complex problems and make them more manageable. Thus, mastering these properties will greatly enhance your algebra skills.
solving equations
Solving equations, particularly those involving logarithms, follows a standard process. First, we strive to isolate the logarithmic part or simplify the given expression using logarithmic properties. Next, we often need to remove the logarithm to solve for the variable.
In our exercise example, after combining the logarithms, we obtained the equation: \(\text{log}_{4} \frac{x+3}{x-1} = 2\).
To solve this, we used the property that if \(\text{log}_{a}(b) = c\), then \(\text{b} = a^{c}\). This enabled us to convert the logarithmic equation to an exponentiation form: \(\frac{x+3}{x-1} = 16\).
Next, we rearranged the equation to solve for \(x\), ensuring each step simplified the equation further, until we reached the solution: \(x = \frac{19}{15}\).
Each step is grounded in the use of basic algebraic principles, making the process systematic and predictable.
In our exercise example, after combining the logarithms, we obtained the equation: \(\text{log}_{4} \frac{x+3}{x-1} = 2\).
To solve this, we used the property that if \(\text{log}_{a}(b) = c\), then \(\text{b} = a^{c}\). This enabled us to convert the logarithmic equation to an exponentiation form: \(\frac{x+3}{x-1} = 16\).
Next, we rearranged the equation to solve for \(x\), ensuring each step simplified the equation further, until we reached the solution: \(x = \frac{19}{15}\).
Each step is grounded in the use of basic algebraic principles, making the process systematic and predictable.
exponentiation
Exponentiation is the process of raising a number (the base) to a power (the exponent). It's a fundamental concept in algebra.
In logarithmic equations, exponentiation often helps undo the logarithm. The core idea here is that logarithms and exponents are inverse operations. For instance, \( \text{log}_{4}(16) = 2 \) because \( 4^2 = 16 \).
In our specific problem, the equation \( \text{log}_{4} \frac{x+3}{x-1} = 2 \) is converted to \( \frac{x+3}{x-1} = 16 \) through exponentiation.
Understanding how to use exponentiation is key to solving equations that involve logarithms. Practice converting between logarithmic and exponential forms to strengthen your mathematical skills. Mastery of this concept will provide a strong foundation for tackling more advanced algebra problems.
In logarithmic equations, exponentiation often helps undo the logarithm. The core idea here is that logarithms and exponents are inverse operations. For instance, \( \text{log}_{4}(16) = 2 \) because \( 4^2 = 16 \).
In our specific problem, the equation \( \text{log}_{4} \frac{x+3}{x-1} = 2 \) is converted to \( \frac{x+3}{x-1} = 16 \) through exponentiation.
Understanding how to use exponentiation is key to solving equations that involve logarithms. Practice converting between logarithmic and exponential forms to strengthen your mathematical skills. Mastery of this concept will provide a strong foundation for tackling more advanced algebra problems.
Other exercises in this chapter
Problem 92
Prove that the area of the triangle with vertices \((0,0),\left(r_{1}, \theta_{1}\right),\) and \(\left(r_{2}, \theta_{2}\right), 0 \leq \theta_{1}
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A 2-pound weight is attached to a 3 -pound weight by a rope that passes over an ideal pulley. The smaller weight hangs vertically, while the larger weight sits
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Use Descartes' Rule of Signs to determine the possible number of positive or negative real zeros for the function $$ f(x)=-2 x^{3}+6 x^{2}-7 x-8 $$
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Find the midpoint of the line segment connecting the points (-3,7) and \(\left(\frac{1}{2}, 2\right)\).
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