Chapter 5: Problem 24
Exer. 11-24: Find the exact solution, using common logarithms, and a two- decimal-place approximation of each solution, when appropriate. $$ 2^{x}-6\left(2^{-x}\right)=6 $$
Short Answer
Expert verified
The exact solution for \(2^x = 3 + \sqrt{15}\) gives \(x \approx 2.77\).
Step by step solution
01
Simplify the Equation
The given equation is \(2^{x} - 6(2^{-x}) = 6\). We can re-write \(2^{-x}\) as \(\frac{1}{2^{x}}\). So the equation becomes:\[2^{x} - \frac{6}{2^{x}} = 6\]To eliminate the fraction, multiply the entire equation by \(2^{x}\):\[(2^x)^2 - 6 = 6 \cdot 2^x\]Thus, the equation transforms to:\[y^2 - 6y - 6 = 0\]where \(y = 2^x\).
02
Use the Quadratic Formula
Now that we have a quadratic equation \(y^2 - 6y - 6 = 0\), we can solve for \(y\) using the quadratic formula. The quadratic formula is given by:\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a = 1\), \(b = -6\), and \(c = -6\). Substitute these into the formula:\[y = \frac{6 \pm \sqrt{36 + 24}}{2}\]This simplifies to:\[y = \frac{6 \pm \sqrt{60}}{2}\]\[y = \frac{6 \pm \sqrt{4 \times 15}}{2}\]\[y = \frac{6 \pm 2\sqrt{15}}{2}\]\[y = 3 \pm \sqrt{15}\]
03
Solve for x
Since \(y = 2^x\), we need to solve \(2^x = 3 + \sqrt{15}\) and \(2^x = 3 - \sqrt{15}\). Calculating the values, we find:\[\sqrt{15} \approx 3.87\]For \(2^x = 3 + \sqrt{15}\), we get \(2^x \approx 6.87\). Taking the logarithm of each side and solving for \(x\):\[\log_{10}(2^x) = \log_{10}(6.87)\]\[x \log_{10}(2) = \log_{10}(6.87)\]\[x \approx \frac{\log_{10}(6.87)}{\log_{10}(2)}\]\[x \approx 2.77\]For \(2^x = 3 - \sqrt{15}\), we find \(3 - \sqrt{15} < 0\), which is not solvable for real numbers since \(2^x\) must be positive.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
The quadratic formula is a reliable method for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). A quadratic equation might not always factorate nicely, making the quadratic formula very handy. The quadratic formula itself is written as:\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
- The term \( \pm \) indicates there are potentially two different solutions for \( y \), depending on whether you add or subtract the square root within the formula.
- \( b^2 - 4ac \) is called the discriminant. It helps determine the nature of the roots:
- if it is positive, the equation has two distinct real roots,
- if it is zero, the equation has exactly one repeated real root,
- if it is negative, the equation has complex roots and no real solutions.
Common Logarithms
Common logarithms are logarithms to the base 10, denoted as \( \log \) or \( \log_{10} \). These are particularly useful because they are directly linked to our standard number system.
- In scientific contexts, the common logarithm is typically used, especially when calculating orders of magnitude or dealing with readable scales like the Richter scale for earthquakes.
- The common logarithm converts multiplicative processes into additive ones, making it easier to manipulate exponential relationships algebraically.
- For example, the property \( \log_{10}(ab) = \log_{10}(a) + \log_{10}(b) \) illustrates how multiplication turns into addition.
Exponential Equations
Exponential equations are equations in which variables appear in exponents. They have the form \( a^{x} = b \), where \( a \) and \( b \) are constants.
- Such equations often model real-world phenomena like population growth, radioactive decay, and interest compounding in finance.
- To solve exponential equations, common strategies include rewriting both sides with a common base or using logarithms.
- Using logarithms is particularly effective as it helps to bring down the variable from the exponent using the property: \( \log_{b}(a^{x}) = x \log_{b}(a) \).