Problem 9 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

In Problems 7 through 32, solve for \(x .\) $$ \pi \cdot 3^{1+2 x}=\sqrt{\pi} 5^{x} $$

Step-by-Step Solution

Answer

The solution to the equation is \(x = \frac{- ln(3) - 0.5 \cdot ln(\pi)}{2 \cdot ln(3) - ln(5)}\).

Step 1: Isolate terms with exponentials on each side

Rewrite the equation as \(3^{1+2x} = \sqrt{\pi}/\pi \cdot 5^x\). This allows us to better work with the exponential terms.

Step 2: Simplify square root

The square root of \(\pi\) divided by \(\pi\) simplifies to \(1/\sqrt{\pi}\). So, the equation becomes \(3^{1+2x} = 1/\sqrt{\pi} \cdot 5^{x}\).

Step 3: Apply logarithms

Taking the natural logarithm (ln) on both sides of the equation will allow us to bring the exponents down. We get \(ln(3^{1+2x}) = ln((1/\sqrt{\pi}) \cdot 5^{x})\). Using the rule \(ln(a \cdot b) = ln(a) + ln(b)\) and the rule \(ln(a^b) = b \cdot ln(a)\), we get \((1+2x) \cdot ln(3) = x \cdot ln(5) - 0.5 \cdot ln(\pi)\).

Step 4: Solve for x

Rearrange terms to solve for \(x\). The equation becomes \((1+2x) \cdot ln(3) - x \cdot ln(5) = -0.5 \cdot ln(\pi)\). Further simplifying gives \(x \cdot (2 \cdot ln(3) - ln(5)) = - ln(3) - 0.5 \cdot ln(\pi)\). Finally, we find \(x = \frac{- ln(3) - 0.5 \cdot ln(\pi)}{2 \cdot ln(3) - ln(5)}\).

Key Concepts

Logarithms

Logarithms are an essential concept to understand when solving exponential equations. They act as the inverse operations of exponentiation. If you picture an exponent as a way to scale a number, logarithms help you find the value that was used as the input. For example, if you know that

• when you raise 2 to a certain power, you get 8, the logarithm helps you determine that power.

In the exercise example, we use natural logarithms, written as \(\ln\), because they are particularly helpful with the constant \(e\), but they can be applied to any base. For instance:

• The natural logarithm of \(e^x\) is simply \(x\).
• When applying logarithms in an equation such as \(\ln(a^b)\), it becomes \(b \cdot \ln(a)\). This property helps us "bring down" any exponents to simplify the calculations.
• Additionally, the rule \(\ln(a \cdot b) = \ln(a) + \ln(b)\) allows breaking down complex terms into manageable pieces.

Solving Equations

When it comes to solving equations, especially those involving exponents and logarithms, it's crucial to isolate the variable you are solving for. Let's consider how to solve an equation step-by-step:

• First, simplify both sides of the equation as much as possible. In our exercise, we rearranged and simplified the terms so that all exponential components are isolated.
• Next, apply logarithms to both sides to transform the exponentiation into multiplication. This makes it easier to move terms around.
• Once you have the terms set out clearly, start rearranging. Group like terms, particularly those containing your variable \(x\), on one side of the equation.
• From there, you can solve for the variable step-by-step with basic algebraic operations, which often involve isolating \(x\) by factoring or simplifying expressions.

In our example, applying these techniques allowed us to express \(x\) explicitly in terms of logarithms.

Exponents

Exponents play a vital role in expressions and equations, signifying repeated multiplication. Yet, when it comes to resolving equations involving exponents, there are some critical steps to bear in mind:

• Understand the base and power: The exponent tells you how many times to multiply the base by itself. For example, \(3^4\) means \(3 \times 3 \times 3 \times 3\).
• In equations, equalizing exponents is an oft-used method to solve for variables. When two exponential expressions with the same base are equal, their exponents can be set equal as well.
• In equations where you have different bases like in the example \(3^{1+2x}\) and \(5^x\), converting to logarithms aids in comparing and solving these terms because it is easier to work with them algebraically once exponents are linear.

In dealing with our specific problem, we used exponent properties for efficient manipulation and solving. Remember the characteristics and rules of exponents as they form the groundwork of manipulating and simplifying expressions with powers.