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Chapter 10: Problem 17

Solve each equation. Use natural logarithms. Approximate solutions to three decimal places when appropriate. $$ e^{0.012 x}=23 $$

Short Answer

Expert verified
x \approx 261.291

Step by step solution

01

Take the natural logarithm on both sides

To solve for the variable, take the natural logarithm (\text{ln}) of both sides of the equation. This helps in dealing with the exponential function. \[\text{ln}\big(e^{0.012x}\big) = \text{ln}(23)\]
02

Apply the property of logarithms

Use the property of logarithms that states \(\text{ln}(e^y) = y\) to simplify the left-hand side of the equation: \[0.012x = \text{ln}(23)\]
03

Solve for x

Isolate the variable x by dividing both sides of the equation by 0.012: \[x = \frac{\text{ln}(23)}{0.012}\]
04

Compute the numerical value

Calculate the value using a calculator. First, find the natural logarithm of 23 and then divide by 0.012. \[x \approx \frac{3.135494}{0.012} \approx 261.291\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Functions
Exponential functions are a type of mathematical function where a constant base is raised to a variable exponent. These functions grow rapidly and are commonly written as \(a^x\). In the exercise, we have the function \(e^{0.012x} = 23\). Here, \(e\) is the base and \(0.012x\) represents the exponent.
Properties of Logarithms
Logarithms are the inverse operations of exponentiation. The natural logarithm, denoted as \(\text{ln}\), follows specific properties that simplify complex equations. One key property used in this exercise is \(\text{ln}(e^y) = y\). This property allows us to take the natural logarithm of both sides and simplify expressions like \(e^{0.012x}\) as \(0.012x\).
Solving Equations
Solving equations involves isolating the variable to find its value. In this exercise, we started with the equation \(e^{0.012x} = 23\). By applying the natural logarithm and the properties of logarithms, we simplified and isolated \(x\):
\[\text{ln}(e^{0.012x}) = \text{ln}(23)\]
Using the property \(\text{ln}(e^y) = y\), the equation simplifies to \(0.012x = \text{ln}(23)\).
Dividing both sides by 0.012 gives us the final solution: \[x \approx \frac{\text{ln}(23)}{0.012} \approx 261.291\].

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Most popular questions from this chapter

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