Problem 24 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 11-24: Find the exact solution, using common logarithms, and a two- decimal-place approximation of each solution, when appropriate. $$ 2^{x}-6\left(2^{-x}\right)=6 $$

Step-by-Step Solution

Answer

The exact solution for \(2^x = 3 + \sqrt{15}\) gives \(x \approx 2.77\).

Step 1: 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 \(y = 2^x\).

Step 2: 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:\[y = \frac{6 \pm \sqrt{60}}{2}\]\[y = \frac{6 \pm \sqrt{4 \times 15}}{2}\]\[y = \frac{6 \pm 2\sqrt{15}}{2}\]\[y = 3 \pm \sqrt{15}\]

Step 3: 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.

Key Concepts

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.

In our exercise, after simplifying the equation, we obtained a quadratic equation \( y^2 - 6y - 6 = 0 \). By identifying \( a = 1 \), \( b = -6 \), and \( c = -6 \), and substituting these into the quadratic formula, we found \( y = 3 \pm \sqrt{15} \). Both solutions were analyzed further to determine feasibility when tackled with the original exponential context.

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.

In the given exercise, after obtaining \( 2^x = 6.87 \), we utilized common logarithms to isolate \( x \). By taking the logarithm of both sides (\( \log_{10}(2^x) = \log_{10}(6.87) \)), we applied the rule \( x \log_{10}(2) = \log_{10}(6.87) \) to solve for \( x \). This yielded the approximation \( x \approx 2.77 \).

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: \( \log_{b}(a^{x}) = x \log_{b}(a) \).

In the problem provided, the exponential equation originally appeared as \( 2^x - 6(2^{-x}) = 6 \). After rewriting and simplifying it further, we used a substitution \( 2^x = y \), transformed it into a quadratic equation, and later reverted to the original format to solve for \( x \). Making use of the logarithmic transformation allowed us to determine if the solutions for \( y \) could yield real and meaningful values for \( x \). This step confirmed that only one of the potential solutions, \( 2^x = 3 + \sqrt{15} \), was valid.