Chapter 5: Problem 76
The number of real roots of the equation \(5+\left|2^{x}-1\right|=2^{x}\left(2^{x}-2\right)\) is: \(\quad\) [April 10,2019 (II)] (a) 3 (b) 2 (c) 4 (d) 1
Short Answer
Expert verified
The number of real roots is 2.
Step by step solution
01
Analyze the Equation
The given equation is:\[ 5 + \left|2^x - 1\right| = 2^x (2^x - 2) \]Start by considering the two cases for the absolute value expression \( \left|2^x - 1\right| \).
02
Case 1: \(2^x - 1 \geq 0\)
Here, \( \left|2^x - 1\right| = 2^x - 1 \). The equation simplifies to:\[ 5 + (2^x - 1) = 2^x (2^x - 2) \]which can be further simplified to:\[ 4 + 2^x = 2^{2x} - 2 \times 2^x \].
03
Simplify Case 1
Rearrange the equation from Step 2:\[ 2^{2x} - 2 \times 2^x - 2^x = 4 \]\[ 2^{2x} - 3 \times 2^x - 4 = 0 \]This is a quadratic equation in terms of \(y = 2^x\):\[ y^2 - 3y - 4 = 0 \].
04
Solve Quadratic for Case 1
Solve the quadratic equation \(y^2 - 3y - 4 = 0\) using the factorization method:\[ (y - 4)(y + 1) = 0 \]Thus, \(y = 4\) or \(y = -1\). Since \(y = 2^x\) and must be positive, only \(y = 4\) is valid, giving \(2^x = 4\) which implies \(x = 2\).
05
Case 2: \(2^x - 1 < 0\)
Here, \( \left|2^x - 1\right| = 1 - 2^x \). The equation becomes:\[ 5 + 1 - 2^x = 2^x (2^x - 2) \]\[ 6 - 2^x = 2^{2x} - 2 \times 2^x \].
06
Simplify Case 2
Rearrange the equation from Step 5:\[ 2^{2x} - 2 \times 2^x + 2^x - 6 = 0 \]\[ 2^{2x} - 1 \times 2^x - 6 = 0 \]This is a quadratic equation in \(y = 2^x\):\[ y^2 - y - 6 = 0 \].
07
Solve Quadratic for Case 2
Solve the quadratic equation \(y^2 - y - 6 = 0\) using factory method:\[ (y - 3)(y + 2) = 0 \]Thus, \(y = 3\) or \(y = -2\). Since \(y = 2^x\) must be positive, only \(y = 3\) is valid, giving \(2^x = 3\) which results in \(x = \log_2(3)\).
08
Verify Cases
Both solutions \(x = 2\) from Case 1 and \(x = \log_2(3)\) from Case 2 satisfy the equation. No other solutions exist as these values are derived comprehensively from the set conditions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Absolute Value Equations
Absolute value equations are a type of mathematical equation involving an absolute value expression. The absolute value of a number represents its distance from zero on a number line, without considering the direction. Hence, the absolute value of a number is always non-negative. For example, \(|x| = 3\) implies either \(x = 3\) or \(x = -3\). When solving absolute value equations, it's crucial to consider both the positive and negative scenarios that could yield the given absolute value.
- Identify the expression within the absolute value.
- Create separate equations for situations where the expression is non-negative and where it is negative.
- Solve each equation separately.
Quadratic Equations
Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. These equations can have two solutions, one solution, or no real solutions depending on the discriminant \(b^2 - 4ac\).
Solving quadratic equations can be done using several methods:
Solving quadratic equations can be done using several methods:
- Factoring the polynomial if possible.
- Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Completing the square to transform the equation into an easier form.
Logarithms
Logarithms are the inverse operations of exponentials. They allow us to solve exponential equations by transforming them into a more manageable form. When you have an equation like \(b^x = y\), it can be rewritten using logarithms as \(x = \log_b(y)\).
The properties of logarithms include:
The properties of logarithms 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: \(\log_b(m^n) = n \cdot \log_b(m)\)