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Magazine Chapter 5: Problem 91
Each of these equations can be transformed into an equation of linear or quadratic type by applying the hint. Solve each equation. (a) \((x-1)^{\log (x-1)}=100(x-1)\) (b) \(\log _{2} x+\log _{4} x+\log _{8} x=11\) (c) \(4^{x}-2^{x+1}=3\)
Short Answer
Expert verified
(a) x = 101; (b) x = 64; (c) x = \log_2 3.
Step by step solution
01
Simplify the equation (a)
First, observe that the equation \((x-1)^{\log(x-1)} = 100(x-1)\) can be simplified. By dividing both sides by \((x-1)\), assuming \(x-1 eq 0\): \[ (x-1)^{\log(x-1)-1} = 100 \]Let \(y = x-1\), then we have:\[ y^{\log(y) - 1} = 100 \]This equation implies \(\log(y) = 2\) because it simplifies into the quadratic-like form \(y^2 = 100,\) once considered under the exponential and logarithmic properties.
02
Solve for y in equation (a)
Since \(\log(y) = 2\) implies \(y = 10^2 = 100,\) substitute back \(y = x - 1\) to find \(x\). This gives us:\[ x - 1 = 100 \]\[ x = 101 \]
03
Solve equation (b)
Convert each term in the equation \(\log_2 x + \log_4 x + \log_8 x = 11\) using the change of base formula:\[ \log_2 x = a, \quad \log_4 x = \frac{a}{2}, \quad \log_8 x = \frac{a}{3} \]This transforms the equation to:\[ a + \frac{a}{2} + \frac{a}{3} = 11 \]Combine the terms over a common denominator:\[ 6a + 3a + 2a = 66 \]\[ 11a = 66 \]
04
Solve for x in equation (b)
Solve for \(a\):\[ a = 6 \]Since \(a = \log_2 x = 6\), it implies:\[ x = 2^6 = 64 \]
05
Transform equation (c)
For the equation \(4^x - 2^{x+1} = 3\), notice that \(4^x\) can be rewritten as \((2^2)^x = (2^x)^2\). Let \(y = 2^x\); then the equation becomes:\[ y^2 - 2y = 3 \]
06
Solve the quadratic equation (c)
Reorganize this into the standard quadratic form:\[ y^2 - 2y - 3 = 0 \]Factoring the quadratic gives:\[ (y - 3)(y + 1) = 0 \]This means the solutions for \(y\) are \(y = 3\) or \(y = -1\). However, since \(y = 2^x\) must be positive, discard \(y = -1\). Hence, \(y = 3\).
07
Solve for x in equation (c)
If \(y = 3\) and \(y = 2^x\), then:\[ 2^x = 3 \]Taking the logarithm of both sides gives us:\[ x = \log_2 3 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Logarithmic Equations
Logarithmic equations involve variables located within a logarithm. Solving them usually requires using logarithmic properties and transforming them into a different form. Let's break down the process.
- Convert logs to a common base: Consider the transformation using the change of base formula. For example, for the terms \(\log_2 x\), \(\log_4 x\), and \(\log_8 x\), convert them to a common base, say base 2.
- Add like terms: Once transformed, add the terms. For instance, \(a + \frac{a}{2} + \frac{a}{3} = 11\) can be solved by finding a common denominator.
- Solve for the variable: After transforming and simplifying, solve for the variable normally, such as finding \(x\) from \(2^x = \text{some value}\).
Quadratic Equations
Quadratic Equations are equations of the form \(ax^2 + bx + c = 0\). They can be solved by a variety of methods including factoring, completing the square, and using the quadratic formula. In our context, they emerged indirectly in transformed forms like \(y^2 - 2y - 3 = 0\).
- Recognize the structure: Understand when an equation can be written in a quadratic form. It might not always be explicit, so identifying patterns can help. For example, converting \(4^x = (2^x)^2\) is key.
- Factor the equation: If it factors neatly like \((y-3)(y+1) = 0\), find the solutions from the factors.
- Validate solutions: Ensure they fit the original variable's conditions. For instance, discard \(y = -1\) since \(2^x\) cannot be negative.
Exponential Functions
Exponential functions take the form \(a^x\), where \(a\) is a constant, and \(x\) is the exponent. They are pivotal when variables appear as exponents.
- Transform for simplification: Convert expressions like \(4^x\) into \((2^2)^x\). Such transformations simplify the algebraic manipulation.
- Utilize logarithms for solving: If stuck with an expression like \(2^x = 3\), apply logarithms to both sides to isolate \(x\). This shifts the problem to a simpler algebraic domain.
- Understand growth and decay: Exponential functions explain rapid growth or decay, which is relevant in real-world applications beyond simple algebra problems.
