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Chapter 6: Problem 4

Simplify the given expression. $$ e^{-2 \ln x} $$

Short Answer

Expert verified
The simplified expression is \( \frac{1}{x^2} \).

Step by step solution

01

Recall the properties of logarithms and exponents

The expression can be simplified using the properties of exponentials and logarithms. Remember, \( e^{a \log b} = b^a \), as long as the logarithm base is equivalent (which is true here with the natural base \( e \)).
02

Use the exponential and logarithmic identity

Apply the identity from the previous step to our expression: \( e^{-2 \ln x} = (e^{\ln x})^{-2} \). Since \( e^{\ln x} = x \), substitute to simplify it to \( x^{-2} \).
03

Interpret the negative exponent

Recall that a negative exponent implies the reciprocal of the base raised to the positive exponent: \( x^{-2} = \frac{1}{x^2} \). Hence, the simplified expression is \( \frac{1}{x^2} \).

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Key Concepts

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

Properties of Exponents
Exponents are mathematical tools that help express repeating multiplications. When it comes to solving expressions, understanding their properties is essential.

**Key Properties**
  • Product of Powers: When you multiply the same bases, you can add the exponents. For example, \( a^b \cdot a^c = a^{b+c} \).
  • Power of a Power: When raising an exponent to another exponent, multiply the exponents. For instance, \( (a^b)^c = a^{bc} \).
  • Power of a Product: When distributing an exponent across multiple bases inside parentheses, each base receives the exponent. This means \( (ab)^c = a^c b^c \).
These properties become vital when dealing with exponential and logarithmic expressions, such as simplifying \( e^{-2 \ln x} \). Knowing how to manipulate exponential identities, like \( e^{a \log b} = b^a \), allows us to transform complex expressions into simpler forms.
Simplifying Expressions
Simplifying mathematical expressions is all about making them easier to work and interpret. For our exercise, the expression was \( e^{-2 \ln x} \).

**Steps To Simplify**
  • First, recognize which properties of exponents and logarithms can be applied. Using the identity \( e^{a \log b} = b^a \) shifts our thinking as it helps simplify expressions involving natural logs.
  • In this case, \( e^{-2 \ln x} \) can be rewritten as \((e^{\ln x})^{-2} \). Since \( e^{\ln x} = x \), it reduces to \( x^{-2} \).
By applying these simplification strategies, complex expressions become manageable. Recognizing and applying the properties appropriately allows you to simplify with confidence and accuracy.
Negative Exponents
Negative exponents signify division instead of multiplication, offering a neat way to express fractions.

**Understanding Negative Exponents**
  • Any expression with a negative exponent, such as \(a^{-b}\), indicates the reciprocal of the base raised to the positive exponent: \( a^{-b} = \frac{1}{a^b} \).
  • In our expression, \( x^{-2} \), this signifies \( \frac{1}{x^2} \).
Negative exponents help make expressions concise, and they're commonly used across various mathematical problems.
In the given problem, understanding that \( x^{-2} \) is simply \( \frac{1}{x^2} \) allowed for a swift and precise final simplification.

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Most popular questions from this chapter

In Problems 13-36, find Dy. $$ y=\operatorname{coth}^{-1}(\tanh x) $$

Let \(f(x)=\cos (\ln x)\). (a) Find the absolute extreme points on [0.1,20] . (b) Find the absolute extreme points on [0.01,20] . (c) Evaluate \(\int_{0.1}^{20} \cos (\ln x) d x\).

Evaluate \(\int_{\pi / 4}^{\pi / 3} \sec x \csc x d x\)

Use the fact that \(\ln 4>1\) to show that \(\ln 4^{m}>m\) for \(m>1 .\) Conclude that \(\ln x\) can be made as large as desired by choosing \(x\) sufficiently large. What does this imply about \(\lim _{x \rightarrow \infty} \ln x ?\)

In Problems 13-36, find Dy. $$ y=\tanh ^{-1}(2 x-3) $$
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