Problem 11 - Contemporary Precalculus

Question

Solve the equation. First express your answer in terms of natural logarithms (for instance, \(x=(2+\ln 5) /(\ln 3)) .\) Then use a calculator to find an approximation for the answer. $$2^{x}=3^{x-1}$$

Step-by-Step Solution

Answer

Question: Solve the equation \(2^x = 3^{x-1}\) and find the approximate value of x. Answer: The solution to the given equation is approximately \(x \approx 1.7095\).

Step 1: Rewrite the equation using natural logarithms

We are given the equation: $$2^{x}=3^{x-1}$$ Rewrite it using natural logarithms by taking ln on both sides: $$\ln{(2^x)} = \ln{(3^{x-1})}$$

Step 2: Apply the properties of logarithms

Now, we will apply the properties of logarithms: Property 1: \(\ln{a^b} = b * \ln{a}\) $$x\ln{2} = (x-1)\ln{3}$$

Step 3: Solve for x,

Rearrange the equation and isolate x: $$x\ln{2} = x\ln{3} - \ln{3}$$ $$x(\ln{2} - \ln{3}) = -\ln{3}$$ Now divide both sides by \((\ln{2} - \ln{3})\) to find the value of x in terms of natural logarithms: $$x = \frac{-\ln{3}}{\ln{2} - \ln{3}}$$

Step 4: Approximate the answer

Now, we will use a calculator to find the numerical approximation of x: $$x \approx \frac{-\ln{3}}{\ln{2} - \ln{3}} \approx 1.7095$$ In conclusion, the solution to the given equation is approximately \(x \approx 1.7095\).

Key Concepts

Properties of Logarithms

Understanding the properties of logarithms is essential when solving equations involving exponents. Logarithms are the opposite of exponential functions, serving as tools that help us decompose complex exponential equations into more manageable linear forms. Here are some key properties of logarithms to keep in mind:

• Product Property: \( \ln{(a \cdot b)} = \ln{a} + \ln{b} \)
• Quotient Property: \( \ln{\left( \frac{a}{b} \right)} = \ln{a} - \ln{b} \)
• Power Property: \( \ln{a^b} = b \cdot \ln{a} \) – This is particularly useful because it allows us to "bring down" exponents for easier manipulation.

In the provided exercise, we used the power property. By re-writing the equation \(2^x = 3^{x-1}\) as \(\ln{(2^x)} = \ln{(3^{x-1})}\), we utilized this property to convert the exponents into coefficients. This transformation made it possible to isolate the variable \(x\) and solve for it efficiently.

Natural Logarithms

Natural logarithms are logarithms with the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. Denoted as \(\ln\), natural logarithms are particularly useful in calculus and continuous growth calculations. Unlike common logarithms, which are based on 10, natural logarithms are natural because they describe growth patterns that occur frequently in nature.

In the context of the equation \(2^x = 3^{x-1}\), taking the natural logarithm of both sides was a strategic step. This approach is beneficial because it allows us to exploit the aforementioned properties of logarithms to simplify and solve the equation. With natural logarithms, manipulation of exponentials into simple linear equations becomes straightforward. Thus, they are a powerful tool in mathematical problem-solving.

• In problems involving exponential functions, rewriting the equation with \(\ln\) can turn multiplication into addition or subtraction, and exponents into coefficients.

Exponential Functions

Exponential functions are characterized by constant growth or decay percentages over time. In mathematics, an exponential function can take the form \(y = a\cdot b^x\). The base \(b\) determines the nature of the exponential growth or decay:

• When \(b > 1\), the function models exponential growth.
• When \(0 < b < 1\), the function models exponential decay.

In the equation \(2^x = 3^{x-1}\), each side represents an exponential function. Here, the bases \(2\) and \(3\) signify exponential growth functions. Solving such equations often requires logarithmic techniques since logarithms are the inverses of exponentials.
By converting the exponential functions into their logarithmic forms, we can employ the properties of logarithms to extract the exponent as a linear function. This simplification is crucial because it allows easier comparison and manipulation of growth rates across different exponential functions, as seen when isolating \(x\) in the provided exercise.