Problem 87 - A Graphical Approach to Precalculus with Limits

Question

The given equations are quadratic in form. Solve each and give exact solutions. $$3^{2 x}+35=12\left(3^{x}\right)$$

Step-by-Step Solution

Answer

Exact solutions are \(x = \log_3{7}\) and \(x = \log_3{5}\).

Step 1: Recognize the Quadratic Form

Notice that the equation \(3^{2x} + 35 = 12 \times 3^x\) is quadratic in form when we substitute \(y = 3^x\). The equation now becomes \(y^2 + 35 = 12y\).

Step 2: Rearrange into Standard Quadratic Equation

Rearrange the equation \(y^2 + 35 = 12y\) into the standard quadratic form \(y^2 - 12y + 35 = 0\).

Step 3: Solve the Quadratic Equation

Factor the quadratic equation \(y^2 - 12y + 35 = 0\). The factors are \((y - 7)(y - 5) = 0\).

Step 4: Find Values of y

Using the factored form, find the values of \(y\) by setting each part equal to zero. So, \(y - 7 = 0\) gives \(y = 7\) and \(y - 5 = 0\) gives \(y = 5\).

Step 5: Substitute Back for x

Recall that \(y = 3^x\). So, for \(y = 7\), we have \(3^x = 7\), leading to \(x = \log_3{7}\). For \(y = 5\), we have \(3^x = 5\), leading to \(x = \log_3{5}\).

Step 6: Express the Exact Solutions

The exact solutions for \(x\) are given by the logarithmic expressions: \(x = \log_3{7}\) and \(x = \log_3{5}\).

Key Concepts

Substitution Method

The substitution method is a powerful technique used to simplify and solve equations, especially when dealing with quadratic or higher-order expressions. At its core, substitution aims to make an equation more manageable by temporarily replacing a complex expression with a simpler variable.

Here's how it works in practice:

• Identify the part of the equation that repeats or is more complex. In this exercise, it's the expression involving exponential terms, specifically \(3^x\).
• Replace this expression with a new variable, such as \(y\). This changes the form of the equation to something simpler, e.g., \(y = 3^x\).
• Solve the new equation, which is typically a more familiar form, like a quadratic equation.
  

The substitution method's beauty lies in its simplicity to handle otherwise difficult expressions by using basic algebraic manipulation.

Factoring Quadratics

Factoring quadratic equations is a technique used to express the quadratic in terms of its roots. This step is crucial for finding solutions easily once the expression is in its simplest factorized form.

Typically, a quadratic expression in the form \(ax^2 + bx + c = 0\) can be factored into the product of two binomials if it satisfies certain criteria:

• Look for two numbers that multiply to give \(ac\) and add to give \(b\).
• Use these numbers to break down the middle term and factor by grouping, if possible.
  

In our case, with \(y^2 - 12y + 35 = 0\), the numbers \(-7\) and \(-5\) satisfy these conditions as \(-7 \times -5 = 35\) and \(-7 + -5 = -12\), allowing it to be written as \((y - 7)(y - 5) = 0\). Correct factoring simplifies solving the equation by isolating each binomial and subsequently finding the roots by setting each factor equal to zero.

Exponential Equations

Exponential equations involve variables in the exponent, making them appear daunting initially. Solving such equations often requires transforming or reinterpreting them using smarter approaches like substitution.

In this scenario, the equation \(3^{2x} + 35 = 12 \times 3^x\) at first looks complex due to the presence of the exponent. By utilizing substitution \((y = 3^x)\), it becomes a conventional quadratic equation \(y^2 + 35 = 12y\), which is simpler to handle.

Remember, the key in dealing with exponential equations is:

• Look for patterns or forms that correspond to simpler expressions.
• Use logarithms if necessary to solve for the variable in the exponent after simplifying the expression.
  

Converting the mathematical challenge by seeing different representations allows solving even intricate exponential equations more effectively.

Logarithms

Logarithms are the inverse operations of exponentiation. They provide a tool for solving equations where the variable is in the power, especially after substitution, as seen in this exercise.

Logarithms translate an exponential equation into a form where the exponent can be isolated and solved. When \(3^x = 7\), using the base-3 logarithm \(x = \log_3{7}\) offers the solution directly.

• The base of the logarithm typically matches the base of the exponent for ease.
• This approach relies on understanding that \(a^b = c\) leads to \(b = \log_a{c}\).
  

Logarithms therefore provide exact solutions and are essential for interpreting exponential equations where direct computation isn't straightforward. Understanding logarithms broadens your mathematical toolset, allowing for the resolution of various types of problems involving growth and decay.