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Magazine Chapter 5: Problem 33
Solve the equation. $$ \log _{3}(x-2)=\log _{3} 27-\log _{3}(x-4)-5^{\log _{5} 1} $$
Short Answer
Expert verified
The solution is \(x = 3 + \sqrt{10}\).
Step by step solution
01
Simplify the Exponent
First, simplify the term with the exponent \(5^{\log _{5} 1}\). Since 5 raised to the power of the logarithm base 5 of 1 is 5鈦�, the expression simplifies to 1. Thus, \(5^{\log _{5} 1} = 1\). Therefore, the given equation changes to: \(\log _{3}(x-2) = \log _{3} 27 - \log _{3}(x-4) - 1\).
02
Apply Logarithmic Properties
Using logarithmic properties, we rewrite the equation. Notice that \(\log _{3} 27 = 3\) since 3鲁 = 27. Now, the equation becomes: \(\log _{3}(x-2) = 3 - \log _{3}(x-4) - 1\). Simplifying further, we get \(\log _{3}(x-2) = 2 - \log _{3}(x-4)\).
03
Combine Logarithms
To combine the logarithms, utilize the property of logarithms: \(\log_b a - \log_b c = \log_b\left(\frac{a}{c}\right)\). Thus, \(\log _{3}(x-2) + \log _{3}(x-4) = 2\) becomes \(\log_3\left((x-2)(x-4)\right) = 2\).
04
Exponentiate to Solve
To remove the logarithm, exponentiate both sides of the equation: \(3^{\log_3\left((x-2)(x-4)\right)} = 3^2\). Thus, \((x-2)(x-4) = 9\).
05
Expand and Solve the Quadratic Equation
Expand the equation \((x-2)(x-4) = 9\) to form a quadratic equation: \(x^2 - 6x + 8 = 9\). Simplify it to \(x^2 - 6x - 1 = 0\) by moving all terms to one side.
06
Solve the Quadratic Equation Using the Quadratic Formula
Apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -6\), and \(c = -1\). Calculate:\[b^2 - 4ac = (-6)^2 - 4 \cdot 1 \cdot (-1) = 36 + 4 = 40\]Thus, \[x = \frac{6 \pm \sqrt{40}}{2}\]Simplifying further,\[x = \frac{6 \pm 2\sqrt{10}}{2}\] Resulting in \(x = 3 \pm \sqrt{10}\).
07
Verify Solutions with Restrictions
We need to check that both solutions satisfy the original equation constraints. Since both must fulfill \(x-2 > 0\) and \(x-4 > 0\), we find:1. \(x = 3 + \sqrt{10}\) is valid as it meets both conditions: \(3 + \sqrt{10} > 4\).2. \(x = 3 - \sqrt{10}\) does not satisfy the requirement since \(3 - \sqrt{10} < 2\). Thus, only \(x = 3 + \sqrt{10}\) is the valid solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Logarithmic Properties
Logarithmic properties are essential tools that help us simplify and solve logarithmic equations. They allow us to manipulate expressions by converting and combining logarithms, making the calculations more manageable.
One vital property is the **power rule**, which states that \( \log_b(a^n) = n \log_b(a) \). This is handy for simplifying expressions where a logarithm is raised to a power.
Another key property is the **product rule**: \( \log_b(m) + \log_b(n) = \log_b(mn) \). This allows you to combine two logarithmic terms into one.
Similarly, the **quotient rule** is used to combine logarithms through division: \( \log_b(m) - \log_b(n) = \log_b \left( \frac{m}{n} \right) \).
In our original exercise, these rules were used to combine logs into a single expression, simplifying the overall process of solving the equation.
One vital property is the **power rule**, which states that \( \log_b(a^n) = n \log_b(a) \). This is handy for simplifying expressions where a logarithm is raised to a power.
Another key property is the **product rule**: \( \log_b(m) + \log_b(n) = \log_b(mn) \). This allows you to combine two logarithmic terms into one.
Similarly, the **quotient rule** is used to combine logarithms through division: \( \log_b(m) - \log_b(n) = \log_b \left( \frac{m}{n} \right) \).
In our original exercise, these rules were used to combine logs into a single expression, simplifying the overall process of solving the equation.
Quadratic Equations
Quadratic equations are polynomials of degree two, usually in the form \( ax^2 + bx + c = 0 \). Solving these equations is an essential part of algebra and can be done using several techniques.
The **Quadratic Formula**, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), provides a way to find the solutions to any quadratic equation. The formula derives from completing the square process and is applicable when factoring isn't simple.
In our problem, after using logarithmic properties to simplify the equation, the result was a quadratic equation. We used the quadratic formula to find potential values for \( x \).
Quadratic equations may have two real solutions, one real solution, or even complex solutions depending on the discriminant \( b^2 - 4ac \). In this exercise, we found two real solutions by evaluating the formula.
The **Quadratic Formula**, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), provides a way to find the solutions to any quadratic equation. The formula derives from completing the square process and is applicable when factoring isn't simple.
In our problem, after using logarithmic properties to simplify the equation, the result was a quadratic equation. We used the quadratic formula to find potential values for \( x \).
Quadratic equations may have two real solutions, one real solution, or even complex solutions depending on the discriminant \( b^2 - 4ac \). In this exercise, we found two real solutions by evaluating the formula.
Solving Equations
To solve equations effectively, you often need to apply multiple mathematical concepts, such as combining different types of equations and leveraging various properties.
The process usually involves simplifying expressions step by step, isolating the variable, or expressing the equation in a more easily solvable form.
In our case, the exercise required starting with logarithmic manipulations, then moving through exponential and polynomial processes to ultimately solve the equation.
Ensuring the validity of solutions is important. Solutions must satisfy the original equation's conditions, such as domain restrictions or basic algebraic requirements.
For example, after solving the quadratic equation part of the exercise, we needed to check that the solutions met the necessary conditions \( x-2 > 0 \) and \( x-4 > 0 \) to validate them in the context of the logarithmic problems.
The process usually involves simplifying expressions step by step, isolating the variable, or expressing the equation in a more easily solvable form.
In our case, the exercise required starting with logarithmic manipulations, then moving through exponential and polynomial processes to ultimately solve the equation.
Ensuring the validity of solutions is important. Solutions must satisfy the original equation's conditions, such as domain restrictions or basic algebraic requirements.
For example, after solving the quadratic equation part of the exercise, we needed to check that the solutions met the necessary conditions \( x-2 > 0 \) and \( x-4 > 0 \) to validate them in the context of the logarithmic problems.
