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Chapter 4: Problem 4

Simplify the following expressions. $$3 \ln \frac{1}{2}+\ln 16$$

Short Answer

Expert verified
\( \text{ln} \, 2 \)

Step by step solution

01

Use the Logarithm Power Rule

Apply the power rule of logarithms to the term \(3 \, \text{ln} \, \frac{1}{2}\). The power rule states that \(a \, \text{ln} \, b = \text{ln} \, b^a\). Therefore, rewrite \(3 \, \text{ln} \, \frac{1}{2}\) as \(\text{ln} \, \left( \frac{1}{2} \right)^3\).
02

Simplify the Exponent

Calculate \( \left( \frac{1}{2} \right)^3\). This equals \( \frac{1}{8}\). So, \( 3 \, \text{ln} \, \frac{1}{2} \) now becomes \( \text{ln} \, \frac{1}{8} \).
03

Use the Logarithm Product Rule

Combine \( \text{ln} \, \frac{1}{8} \) and \( \text{ln} \, 16 \) using the logarithm product rule which states that \( \text{ln} \, a + \text{ln} \, b = \text{ln} \, (a \cdot b) \). This yields \( \text{ln} \, \left( \frac{1}{8} \cdot 16 \right) \).
04

Simplify the Product

Calculate \( \frac{1}{8} \cdot 16 \). This simplifies to 2. Therefore, \( \text{ln} \, \left( \frac{1}{8} \cdot 16 \right) = \text{ln} \, 2 \).

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Key Concepts

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

Logarithm Power Rule
The logarithm power rule is a key tool in simplifying logarithms. This rule states that for any constants \(a\) and \(b\), we have \(a \ln b = \ln b^a\).
Basically, if a logarithm is multiplied by a number, you can move this number as an exponent inside the logarithm. For example, \(3 \ln \left( \frac{1}{2} \right)\) can be rewritten as \( \ln \left( \left( \frac{1}{2} \right)^3 \right) \).
This transforms our initial expression and simplifies further steps. Remember, the power rule is useful when you have a log term multiplied by a number.
Logarithm Product Rule
The logarithm product rule is another vital concept in simplifying logarithms. It states that \( \ln a + \ln b = \ln (a \cdot b) \).
By using this rule, we can combine two separate logarithms into one.
For example, after rewriting \(3 \ln \left( \frac{1}{2} \right)\) as \( \ln \left( \frac{1}{8} \right)\), we can combine it with \( \ln 16 \)
as \( \ln \left( \left( \frac{1}{8} \right) \cdot 16 \right) \).
This rule is particularly useful for adding logs together, simplifying the overall expression.
Simplifying Logarithms
Simplifying logarithms involves using the rules of logarithms to break down complex expressions into simpler, more manageable ones.
Different rules, like the power rule and product rule, are used in conjunction for this purpose.
Let's walk through our example:
  • First, we applied the power rule: \( 3 \ln \left( \frac{1}{2} \right)\) became \( \ln \left( \left( \frac{1}{2} \right)^3 \right) \), which simplifies to \( \ln \left( \frac{1}{8} \right) \).

  • Next, we used the product rule to combine \( \ln \left( \frac{1}{8} \right)\) and \( \ln 16 \): \( \ln \left( \frac{1}{8} \cdot 16 \right) \), simplifying to \( \ln 2 \).
Understanding these rules and how to apply them step-by-step is the foundation for mastering logarithms.

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Most popular questions from this chapter

Differentiate. $$y=\ln \frac{x+1}{x-1}$$

Which of the following is the same as \(\ln (9 x)-\ln (3 x) ?\) (a) \(\ln 6 x\) (b) \(\ln (9 x) / \ln (3 x)\) (c) \(6 \cdot \ln (x)\) (d) \(\ln 3\)

Find the values of \(x\) at which the function has a possible relative maximum or minimum point. (Recall that \(e^{x}\) is positive for all \(x .\) ) Use the second derivative to determine the nature of the function at these points. $$f(x)=\frac{4 x-1}{e^{x / 2}}$$

(a) Use the fact that \(e^{4 x}=\left(e^{x}\right)^{4}\) to find \(\frac{d}{d x}\left(e^{4 x}\right) .\) Simplify the derivative as much as possible. (b) Take an approach similar to the one in (a) and show that, if \(k\) is a constant, \(\frac{d}{d x}\left(e^{k x}\right)=k e^{k x}.\)

Differentiate the following functions. $$y=\left(x^{2}+x+1\right) e^{x}$$
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