Problem 88 - A Graphical Approach to College Algebra

Question

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

Step-by-Step Solution

Answer

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

Step 1: Recognize Substitution Form

The equation \(5^{2x} + 3(5^x) = 28\) can be transformed by setting \(5^x = y\). This allows the equation to be rewritten as \(y^2 + 3y = 28\).

Step 2: Rearrange into a Quadratic Equation

Rearrange the equation into the standard quadratic form \(y^2 + 3y - 28 = 0\).

Step 3: Solve the Quadratic Equation

Use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = 3\), \(c = -28\). Calculate the discriminant: \(b^2 - 4ac = 3^2 - 4(1)(-28) = 121\).

Step 4: Calculate Quadratic Solutions

Substitute back into the quadratic formula: \(y = \frac{-3 \pm \sqrt{121}}{2}\). This simplifies to \(y = \frac{-3 \pm 11}{2}\), yielding \(y = 4\) and \(y = -7\).

Step 5: Back Substitute for Original Variable

Recall that \(y = 5^x\). Thus, \(5^x = 4\) for the first value. The solution \(5^x = -7\) is not possible since exponential functions are always positive.

Step 6: Solve for x

Solve \(5^x = 4\) by taking the logarithm base 5 of both sides: \(x = \log_5 4\). Using the change of base formula, this becomes \(x = \frac{\log 4}{\log 5}\).

Key Concepts

Exponential Equations

Exponential equations involve variables that appear as exponents. In these equations, such as the problem we are tackling, variables are in the form of powers of a given base. For instance, in the equation \(5^{2x} + 3(5^x) = 28\), the variable \(x\) is an exponent for the base 5. Solving exponential equations typically involves expressing them in a way that allows for easier manipulation, such as factoring or rewriting using substitution.

Here are steps generally followed to solve exponential equations:

• Identify similar exponential terms in the equation.
• Use substitutions if the equation is quadratic in form to simplify it.
• Once simplified, additional methods like the quadratic formula or logarithms can be applied.

Usually, once the exponential terms are managed, the solution involves back-transforming the substitution to return to the original base/exponent format. This ensures the solution captures the original structure of the problem.

Change of Base Formula

The change of base formula is an essential tool in solving logarithmic equations, especially when a calculator is needed and doesn't support different logarithm bases directly. It allows one to rewrite logarithms from one base to another. This becomes handy when solving equations like \(x = \log_5 4\) where most calculators only have natural logs (base \(e\)) and common logs (base 10).The change of base formula is stated as follows:\[\log_b a = \frac{\log_c a}{\log_c b}\]Typically, we choose base \(c\) to be either 10 or \(e\) because these are the bases available on most calculators:

• \(\log a = \log_{10} a\) for common logarithms
• \(\ln a = \log_{e} a\) for natural logarithms

This formula allows us to represent a logarithm in a different base using the ones available on calculators, making calculations practical and efficient.

Quadratic Formula

The quadratic formula is a consistent tool used to solve quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). In our exercise, we derived to such a form, \(y^2 + 3y - 28 = 0\), after substitution. The solution involves using the quadratic formula:\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here's the application to our exercise:

• Plugging \(a = 1\), \(b = 3\), \(c = -28\) into the formula.
• Calculating the discriminant \(b^2 - 4ac\) which, when positive, indicates real solutions.
• Solving yields \(y = 4\) and \(y = -7\) but only \(y = 4\) is valid under our original setup.

The quadratic formula is a reliable method because it always delivers results when real solutions exist. It’s a critical step in finding the exact solutions required in mathematical problems.

Logarithms

Logarithms are the inverse operation to exponentiation and are incredibly useful in solving equations where the variable is an exponent. Essentially, solving the equation \(5^x = 4\) involves "undoing" the exponentiation. We apply the logarithm to both sides of the equation to isolate the variable \(x\):\[x = \log_5 4\]In this process, the change of base formula might be used to calculate \(\log_5 4\) should a calculator need it:\[x = \frac{\log 4}{\log 5}\]
Logarithms help us transform multiplicative relationships into additive ones, making equations and calculations more manageable:

• \(\log(ab) = \log a + \log b\) turns multiplication into addition.
• \(\log(a^b) = b \log a\) brings exponents down.

Understanding how to manipulate equations using logarithms is an invaluable step in solving complex exponential forms, just as seen in this exercise.