Problem 8 - Precalculus with Calculus Previews

Question

Solve the given exponential equation. $$ \frac{4^{x}}{3^{x}}=\frac{9}{16} $$

Step-by-Step Solution

Answer

The solution is approximately \(x = 2\).

Step 1: Convert the Bases

The equation given is \( \frac{4^x}{3^x} = \frac{9}{16} \). Notice that \(4\) and \(16\) can be expressed as powers of \(2\), and \(3\) and \(9\) can be expressed as powers of \(3\). So, rewrite the equation using these expressions: \( \frac{(2^2)^x}{(3^1)^x} = \frac{(3^2)}{(2^4)} \). This simplifies the equation to \( \frac{2^{2x}}{3^x} = \frac{3^2}{2^4} \).

Step 2: Equate the Exponents

Since the bases are the same, the exponents can be equated. Rewrite the equation as \( 2^{2x} \cdot 2^4 = 3^2 \cdot 3^x \), which gives us \( 2^{2x+4} = 3^{x+2} \). Thus we have two separate exponent equations: \( 2x + 4 \) for the \(2\) exponent and \(x + 2\) for the \(3\) exponent.

Step 3: Solve for x Using Logarithms

Now we need to solve for \(x\). Using logarithms, take the natural log of both sides: \( \ln(2^{2x+4}) = \ln(3^{x+2}) \). This simplifies by the power rule of logarithms: \( (2x + 4) \ln(2) = (x + 2) \ln(3) \).

Step 4: Isolate x

Rearrange the equation to solve for \(x\): \( 2x \ln(2) + 4 \ln(2) = x \ln(3) + 2 \ln(3) \). Now, bring all the \(x\) terms to one side: \( 2x \ln(2) - x \ln(3) = 2 \ln(3) - 4 \ln(2) \). Factor out \(x\): \( x (2 \ln(2) - \ln(3)) = 2 \ln(3) - 4 \ln(2) \).

Step 5: Solve for x Numerically

Solve for \(x\) by dividing each side of the equation: \( x = \frac{2 \ln(3) - 4 \ln(2)}{2 \ln(2) - \ln(3)} \). Simplify this fraction to find the numerical value of \(x\).

Key Concepts

Logarithms

Logarithms are a powerful mathematical tool used to solve equations involving exponents. They help us "bring down" the exponents to a level where we can easily manage and manipulate them. For instance, if we have an equation like \(a^x = b\), it can be difficult to directly solve for \(x\), especially if the bases are not easily comparable. By taking the logarithm of both sides, we convert the problem into a simpler form: \(x \cdot \log(a) = \log(b)\). This transformation underlines one of the key properties of logarithms - they convert multiplication into addition, which is often easier to work with.

• This was used in the solution when we took the natural log of both sides: \(\ln(2^{2x+4}) = \ln(3^{x+2})\).
• Using properties of logarithms like \(\ln(a^b) = b \cdot \ln(a)\), the equation simplified to \((2x+4) \ln(2) = (x+2) \ln(3)\).

Base Conversion

Base conversion helps simplify equations with different bases into a comprehensible form. If an equation involves exponential terms with different bases, it can be re-expressed using a common base or a simpler power expression. For example, in the problem \(\frac{4^x}{3^x} = \frac{9}{16}\), we noted that both \(4\) and \(16\) can be expressed in terms of base \(2\) (\(4 = 2^2\) and \(16 = 2^4\)), and \(3\) and \(9\) can be expressed in terms of base \(3\) (\(9 = 3^2\)).

• By recognizing these relationships, we rewrote the given equation as \(\frac{(2^2)^x}{(3^1)^x} = \frac{3^2}{2^4}\).
• This conversion made it easier to manage the expressions and served as a foundation for the next step: equalizing exponents.

Base conversion is particularly useful when dealing with standardized tests or exercises that frequently incorporate such transformations, as it aids in pattern recognition.

Solving Equations

Solving equations typically involves finding the unknown values that satisfy the equation. In the exercise, the unknown was \(x\), and we needed to find its value that makes the equation true. The approach involved several strategic steps:

• Firstly, base conversion was used to rewrite the equation in manageable terms \(2^{2x} \text{ and } 3^{x}\).
• After expressing the equation with uniform bases, the step "equate the exponents" allowed us to focus on the terms corresponding to common bases.
• Finally, logarithms were used to convert the equation into a linear format from an exponential one.

Isolating the variable "\(x\)" often requires rearranging terms, using identities, and involving arithmetic operations until \(x\) is isolated. In the solution, by bringing all terms involving \(x\) to one side, and relying on logarithm properties, the equation was distilled into a solvable form.

Exponents

Exponents signify repeated multiplication. They play a critical part in expressing large numbers or making calculations simpler. In this equation, multiple terms had exponents like \(4^x\) and \(3^x\), which were simplified by leveraging exponent rules.

• The expression \(4^x\) was rewritten as \((2^2)^x\), which became \(2^{2x}\) via the exponent multiplication rule: \((a^m)^n = a^{m \cdot n}\).
• Similarly, \(3^x\) stayed as \((3^1)^x\), maintaining consistency in the base conversion process.

The transformation of these terms into a common base facilitates an intuitive understanding of the manipulation process leading down through the solution. Understanding how exponents work and their properties like "product and quotient rule" and "power of a power rule" often aids in breaking down complex exponential equations into simpler components, much like unraveling multiple layers of a complex problem.