Problem 29 - Contemporary Precalculus

Question

Solve the equation for \(x\) by first making an appropriate substitution. $$4^{x}+6 \cdot 4^{-x}=5$$

Step-by-Step Solution

Answer

Answer: The two solutions for \(x\) are: 1. \(x = \log_4{2}\) 2. \(x = \log_4{3}\)

Step 1: Make a substitution

Let's make a substitution to simplify this equation. Let \(y = 4^x\). Then, the reciprocal of \(y\) is \(1/y\). We can write the equation as follows: $$y + 6\cdot\frac{1}{y} = 5$$

Step 2: Clear the fraction and rearrange the terms

To clear the fraction, we can multiply both sides of the equation by \(y\): $$y^2 + 6 = 5y$$ Now, let's rearrange this equation into the standard form of a quadratic equation: $$y^2 - 5y + 6 = 0$$

Step 3: Solve the quadratic equation

We can now solve the quadratic equation either by factoring, completing the square, or using the quadratic formula. We will use factoring here. We can factor the equation like this: $$(y-2)(y-3) = 0$$ Set each factor equal to zero and solve for \(y\): $$y-2 = 0 \Rightarrow y = 2$$ $$y-3 = 0 \Rightarrow y = 3$$

Step 4: Reverse the substitution and solve for x

Now, let's reverse the substitution made in step 1 and use the values of \(y\) obtained in step 3 to find \(x\). Recall that \(y = 4^x\): $$4^x = 2 \Rightarrow x = \log_4{2}$$ $$4^x = 3 \Rightarrow x = \log_4{3}$$ The two solutions for \(x\) are: $$x = \log_4{2}$$ $$x = \log_4{3}$$

Key Concepts

Substitution Method

The substitution method is a useful algebraic technique to solve equations, especially when dealing with complex expressions. The main goal is to simplify the original equation by introducing a new variable, which transforms the equation into a more manageable form. In the given problem, we start by substituting the expression involving exponents, \(4^x\), with a new variable \(y\). This change alters the original exponential equation:\[4^{x} + 6 \cdot 4^{-x} = 5\]into a simpler algebraic equation:\[y + 6 \cdot \frac{1}{y} = 5\] This substitution reduces the complexity of dealing with exponents directly. It allows us to solve the equation in terms of the new variable \(y\). After solving for \(y\), we reverse the substitution to find the original variable, \(x\). This method is especially helpful for equations involving reciprocal or inverse functions.

Quadratic Equation Solution

After substitution, the problem becomes a quadratic equation, \(y^2 - 5y + 6 = 0\). Quadratic equations can be solved using several methods:

• Factoring: This involves expressing the quadratic as a product of two binomials. For \(y^2 - 5y + 6\), it factors nicely into \((y-2)(y-3) = 0\).
• Quadratic Formula: For \(ax^2+bx+c=0\), the solution is given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Use when factoring is complex.
• Completing the Square: Another method if the quadratic doesn't factor neatly or to derive the quadratic formula.

In this problem, factoring was straightforward, and we obtained solutions \(y = 2\) and \(y = 3\). Lastly, with the solutions for \(y\) in hand, we proceed to find the value of the initial variable.

Logarithmic Equations

Upon solving the quadratic equation in terms of \(y\), we go back to the original terms by analyzing \(y = 4^x\). We discover two equations: \(4^x = 2\) and \(4^x = 3\). To find \(x\), we need to express these equations logarithmically.To solve \(4^x = 2\) using logarithms, we write:\[x = \log_4{2}\]This expression means "what power should 4 be raised to, to result in 2?" Similarly, for \(4^x = 3\):\[x = \log_4{3}\]Logarithms help us find the exponent solution to exponential forms when they aren't immediately apparent. The use of logarithms is vital for solving equations where the variable is an exponent. It provides access to otherwise difficult exponential calculations, making this problem easier to comprehend.