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Chapter 12: Problem 61

Evaluate each expression without using a calculator. $$\ln e^{6}$$

Short Answer

Expert verified
The evaluation of \( \ln e^{6} \) without using a calculator results in 6.

Step by step solution

01

Recognize and apply logarithmic property

The given expression is \( \ln e^{6} \). We can use the property of logarithms that says the logarithm of a number raised to an exponent is equal to the exponent times the logarithm of the number. Hence, \( \ln e^{6} \) becomes \( 6 \cdot \ln e \).
02

Evaluate the logarithm of e

We know that \( \ln e =1 \) because e is the base of the natural logarithm. So, \( 6 \cdot \ln e \) becomes \( 6 \cdot 1 \).
03

Calculate the final result

Finally, multiply 6 by 1 to obtain the result, which is 6.

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Key Concepts

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

Exponentiation
Exponentiation is a mathematical operation involving two numbers, where one number (the base) is raised to the power of another number (the exponent). The expression \( e^6 \) consists of the base \( e \), the mathematical constant approximately equal to 2.718, raised to the 6th power. This operation implies multiplying \( e \) by itself six times.

Here are some important points about exponentiation:
  • Exponentiation is repeated multiplication. For example, \( e^6 = e \times e \times e \times e \times e \times e \).
  • The base \( e \) is unique because it is the base of the natural logarithm, which appears frequently in calculus and complex number theory.
  • Understanding exponentiation is crucial for simplifying expressions involving powers, as shown in the problem \( \ln e^6 \).
Through exponentiation, complex expressions become manageable, especially when combined with logarithmic properties.
Logarithmic Properties
Logarithms are the inverse operations of exponentials, allowing us to "undo" exponentiation. For natural logarithms (logarithms with base \( e \)), the notation is \( \ln \, \).

Let's break down the key properties:
  • The Power Rule of logarithms, used in this exercise, states that \( \ln (a^b) = b \cdot \ln a \). This property simplifies expressions where numbers are raised to a power before applying the logarithm.
  • So, \( \ln e^6 \) simplifies to \( 6 \cdot \ln e \) using this rule, turning a complex expression into a manageable product.
  • The natural log of \( e \), \( \ln e \), is always 1. This is because \( e \) is the base of its own logarithmic function, which simplifies calculations.
Applying these properties efficiently allows one to simplify complex logarithmic expressions swiftly, which is especially helpful when solving problems without a calculator.
Evaluating Expressions
To evaluate expressions like \( \ln e^6 \), understanding the underlying properties is essential. Evaluation here involves using known properties to simplify and solve expressions without computational tools.

In the exercise:
  • Start by recognizing the form of \( \ln e^6 \). Using the Power Rule of Logarithms, rewrite it: \( \ln e^6 = 6 \cdot \ln e \).
  • By knowing that \( \ln e = 1 \), substitute and simplify the expression. The evaluation becomes \( 6 \cdot 1 = 6 \).
These steps clearly show the relationship between exponentials and logarithms in evaluation. The evaluation process displays the power of leveraging mathematical properties to obtain results efficiently, emphasizing the importance of core mathematical insights.

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Most popular questions from this chapter

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. Examples of exponential equations include \(10^{x}=5.71\) \(e^{x}=0.72,\) and \(x^{10}=5.71\)

Simplify: \(\left(-2 x^{3} y^{-2}\right)^{-4}\)

Solve each equation. $$3^{x+2} \cdot 3^{x}=81$$

Solve: $$\frac{3}{x+1}-\frac{5}{x}=\frac{19}{x^{2}+x}$$

Will help you prepare for the material covered in the next section. In each exercise, evaluate the indicated logarithmic expressions without using a calculator. a. Evaluate: \(\log _{2} 32\) b. Evaluate: \(\log _{2} 8+\log _{2} 4\) c. What can you conclude about \(\log _{2} 32,\) or \(\log _{2}(8 \cdot 4) ?\)
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