Problem 33 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

In Problems 33 throuogh 36, solve for \(x ; O, R\), and \(S\) are positive constants. (a) \(3^{5 x+2}=100\) (b) \(Q^{2 x+1}=R\)

Step-by-Step Solution

Answer

The solution to (a) is \(x \approx -0.25\) and the solution to (b) is \(x = \frac{\log_{Q} (R) - 1}{2}\).

Step 1: Rearrange the first equation

Rewrite equation \(3^{5x+2} = 100\) as \(5x + 2 = \log_{3} (100)\). This step relies on the fact that \(a = b^c\) can be rewritten as \(c = \log_{b}(a)\).

Step 2: Solve for \(x\)

Subtract 2 from both sides followed by dividing by 5. This results in: \(x = \frac{\log_{3} (100) - 2}{5}\). Calculating \(\log_{3} (100)\), we find that \(x \approx -0.25\).

Step 3: Rearrange the second equation

Rewrite equation \(Q^{2x+1} = R\) as \(2x + 1 = \log_{Q} (R)\). This step again uses the property \(a = b^c \rightarrow c = \log_{b}(a)\).

Step 4: Solve for \(x\)

Subtract 1 from both sides and then divide by 2. Therefore, \(x = \frac{\log_{Q} (R) - 1}{2}\).

Key Concepts

Logarithms

Logarithms are incredibly useful in solving exponential equations. Think of logarithms as the opposite of exponentiation, just like subtraction is the opposite of addition. If you're trying to solve an equation like \(a = b^c\), where \(b\) is the base, \(c\) is the exponent, and \(a\) is the result, you can transform it into \(c = \log_{b}(a)\). This transformation helps in isolating the variable you're solving for, especially when it's in the form of an exponent in an equation.

To put it simply, the logarithm answers the question, "To what power must the base be raised, to yield a certain number?" This concept is crucial when dealing with exponential equations as logarithms simplify the complex task of taking out the variable from an exponential form.

Solving Equations

Solving equations, especially those involving exponents, often requires a methodical approach. In the given problems, we start by expressing the equation in a form where the variable can be isolated. For example, given the equation \(3^{5x+2} = 100\), our goal is to find the value of \(x\).

The first step involves rewriting the equation in logarithmic form: \(5x + 2 = \log_{3}(100)\). This step converts the exponential equation into a linear one, making it simpler to solve. From here, solving for \(x\) becomes a matter of simplifying: subtract \(2\) from both sides and then divide by \(5\). These operations allow us to isolate \(x\), hence solving the equation effectively.

Remember, solving equations is about applying the correct algebraic manipulations to both sides of the equation until the variable is isolated.

Algebraic Manipulation

Algebraic manipulation is the process of using algebraic techniques to rearrange and simplify equations. Each manipulation step is crucial when solving equations, particularly those containing variables in the exponent.

In exponential equations, like \(Q^{2x+1} = R\), our goal is to isolate the variable by making the equation simpler. Once rewritten in logarithmic form: \(2x + 1 = \log_{Q}(R)\), we perform algebraic manipulations. By subtracting \(1\) from both sides, and then dividing everything by \(2\), we isolate \(x\) — the desired quantity.

Algebraic manipulation involves stepping through various arithmetic steps like addition, subtraction, multiplication, or division that maintains the equality of the equation. These steps allow us to peel away layers of complexity, revealing the variable we seek to solve.