Problem 88 - A Graphical Approach to Precalculus with Limits

Question

The given equations are quadratic in form. Solve each and give exact solutions. $$5^{2 x}+3\left(5^{x}\right)=28$$

Step-by-Step Solution

Answer

The exact solution is \( x = \frac{\log(4)}{\log(5)} \).

Step 1: Identify the Substitution

Notice that the equation \(5^{2x} + 3 \times 5^x = 28\) can be rewritten using substitution. Let \( y = 5^x \). Then, \( 5^{2x} = (5^x)^2 = y^2 \). The equation becomes \( y^2 + 3y = 28 \).

Step 2: Rearrange the Equation

Rearrange the equation obtained from the substitution to standard quadratic form: \( y^2 + 3y - 28 = 0 \). This is now a quadratic equation in terms of \( y \).

Step 3: Solve the Quadratic Equation

Use the quadratic formula to solve for \( y \): \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 3 \), \( c = -28 \). Substitute these values to get \( y = \frac{-3 \pm \sqrt{3^2 - 4 \times 1 \times (-28)}}{2 \times 1} \).

Step 4: Simplify the Solution

Calculate the discriminant: \( 3^2 - 4 \times 1 \times (-28) = 9 + 112 = 121 \). This is a perfect square. Simplify to solve for \( y \): \( y = \frac{-3 \pm 11}{2} \). This gives solutions \( y = 4 \) and \( y = -14 \). Since \( y = 5^x \) must be positive, we take \( y = 4 \).

Step 5: Find the Value of x

Since \( y = 5^x \), solve \( 5^x = 4 \). Take the logarithm of both sides (base 10 or base \( e \)): \( x \log(5) = \log(4) \). Thus, \( x = \frac{\log(4)}{\log(5)} \).

Step 6: Final Solution and Verification

Calculate \( x \approx \frac{0.602}{0.699} \approx 0.862 \). Verify by substituting \( x \approx 0.862 \) back into the original equation to confirm that it satisfies \( 5^{2x} + 3 \times 5^x = 28 \).

Key Concepts

Substitution Method

The substitution method is a valuable solving technique used for equations that can be cumbersome to tackle in their original form, particularly in polynomial equations. In the given problem, the equation features exponential terms, specifically with the base of 5 raised to powers involving the variable \(x\). A clever way to simplify this is to substitute a new variable for the exponential term. For example, let \(y = 5^x\). This transforms \(5^{2x}\) into \((5^x)^2\), which can be rewritten as \(y^2\). Now, the equation changes from \(5^{2x} + 3 \times 5^x = 28\) to \(y^2 + 3y = 28\).
By converting the equation into an easier-to-solve form, we can bring it down to a standard quadratic equation. We then solve the quadratic equation for \(y\) and subsequently find the values of \(x\) using the relationship \(y = 5^x\). This method can dramatically simplify solving complex equations, revealing an underlying structure that is much more manageable.

Quadratic Formula

Quadratic equations are often solved using the quadratic formula, which provides a systematic way to find the solutions, or 'roots', of the equation. The quadratic formula is written as:

• \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, \(a\), \(b\), and \(c\) are coefficients from the standard form of a quadratic equation: \(ax^2 + bx + c = 0\). In this exercise, after substituting \(y = 5^x\), the equation becomes \(y^2 + 3y - 28 = 0\). Thus, \(a = 1\), \(b = 3\), \(c = -28\).
In applying the quadratic formula, the discriminant \(b^2 - 4ac\) plays a crucial role as it determines the nature of the roots:

• If the discriminant is positive, there are two distinct real roots.
• If it's zero, there's exactly one real root.
• If negative, the roots are complex and not real numbers.

For our equation, the discriminant is \(3^2 - 4 \times 1 \times (-28) = 121\), which is positive and a perfect square. Hence, the equation has two real solutions: \(y = 4\) and \(y = -14\). However, only \(y = 4\) is valid for \(5^x = y\), as \(5^x\) must be a positive number.

Solving Exponential Equations

Exponential equations involve variables in the exponents, such as \(5^x = 4\) in our final step after the substitution process. To solve such equations, logarithms are typically used to isolate the variable. Logarithms act as the inverse operation to exponentiation. In the case where \(5^x = 4\), taking the logarithm of both sides allows us to bring the variable \(x\) down from the exponent:

• \(x \log(5) = \log(4)\)

Solving for \(x\) involves dividing both sides by \(\log(5)\), leading to:

• \(x = \frac{\log(4)}{\log(5)}\)

This gives an approximate decimal solution that can be calculated numerically. The process of solving such exponential equations by taking logarithms is straightforward once the equation is appropriately set up. It enables us to effectively find the solutions even when the base and the exponent make calculations non-trivial. Understanding this method is crucial whenever dealing with equations that incorporate exponential terms.