Problem 76 - College Algebra with Modeling and Visualization

Question

Solve each equation. Use the change of base formula to approximate exact answers to the nearest hundredth when appropriate. $$2(3)^{-2 x}+5=167$$

Step-by-Step Solution

Answer

The solution to the equation is \(x = -2\).

Step 1: Isolate the Exponential Expression

Start with the equation \(2(3)^{-2x} + 5 = 167\). First, subtract 5 from both sides to isolate the exponential term: \[ 2(3)^{-2x} = 162.\]

Step 2: Divide Both Sides by 2

Next, divide both sides of the equation by 2 to simplify it further: \[ (3)^{-2x} = 81.\]

Step 3: Rewrite as a Logarithm

Recognize that \((3)^{-2x} = 3^{-2x} = 81\). Turn this exponential equation into a logarithmic equation using the change of base formula: \[ -2x = \log_3{81}.\]

Step 4: Evaluate the Logarithm

Calculate \(\log_3{81}\). Since \(81 = 3^4\), it follows that \(\log_3{81} = 4\). Thus, we have: \[ -2x = 4.\]

Step 5: Solve for x

Now, solve for \(x\) by dividing both sides of the equation by -2: \[ x = \frac{4}{-2} = -2.\]

Key Concepts

Exponential Equations

Exponential equations are equations in which a variable appears in the exponent. This type of equation is crucial in various fields, such as compound interest calculations and growth models. In our example, we have an equation of the form \( 2(3)^{-2x} + 5 = 167 \). Here, the exponential part is \((3)^{-2x}\). This represents a rapid change, and solving such equations often involves isolating the exponential expression first.
When tackling exponential equations, you'll often need to:

• Isolate the exponential part by removing any constants added or multiplied to it.
• Rewrite the equation to make it easier to solve using logarithms or base conversion techniques.

Solving these equations typically requires several steps, including algebraic manipulation and sometimes logarithmic conversion when the exponent contains the variable.

Logarithmic Conversion

Logarithmic conversion is a technique commonly used to turn an exponential equation into a form that is simpler to solve. This method uses logs to "undo" the exponentiation. In our solved example, after isolating the exponential term \((3)^{-2x} = 81\), we convert this exponential equation into a logarithmic form using the change of base formula.
The change of base formula offers a way to compute logarithms by converting them to a different base. Specifically, for our task, converting the expression \( -2x = \log_3{81} \) utilizes the fact that if \( b^y = x \), then \( y = \log_b{x} \). Since \(81\) is \(3^4\), it follows that \( \log_3{81} = 4 \).
Key points to remember when using logarithmic conversion:

• Recognize the exponential form of the equation and rewrite it as a logarithm.
• Evaluate or simplify log expressions where possible, especially when they have known values like powers of a base.

This method leverages the properties of logarithms to solve for the variable inside the exponent, making complex calculations more manageable.

Algebraic Manipulation

Algebraic manipulation involves re-arranging an equation to isolate terms and simplify calculations. This is essential both before and after converting the equation into a logarithmic form. In the given problem, algebraic steps play a significant role from start to finish.
Initially, we needed to isolate the exponential expression by removing constants on the same side of the equation, starting with:

• Subtracting 5 from both sides to remove the additional term: \(2(3)^{-2x} + 5 = 167\) becomes \(2(3)^{-2x} = 162\).
• Dividing by 2 to simplify: \((3)^{-2x} = 81\).

After simplifying the equation into a logarithmic form \( -2x = \log_3{81} \), algebraic manipulation again helps find the precise solution:

• Solve \(-2x = 4\) by dividing both sides by \(-2\), yielding \(x = -2\).

Mastering algebraic manipulation is vital because it helps in breaking down complex problems, making them easier to handle with straightforward operations.