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91Ó°ÊÓ

Use the properties of natural logarithms to simplify each function. $$ f(x)=8 x-e^{\ln x} $$

Short Answer

Expert verified
The simplified function is \( f(x) = 7x \).

Step by step solution

01

Understanding the Problem

We have the function \( f(x) = 8x - e^{\ln x} \). We need to simplify this function using the properties of natural logarithms.
02

Applying the Exponential and Logarithm Relationship

Recall that the exponential and natural logarithm functions are inverse operations, so \( e^{\ln x} = x \). This is a fundamental property of natural logarithms and exponentials.
03

Substitute and Simplify

Using the property \( e^{\ln x} = x \) from the previous step, substitute \( x \) in place of \( e^{\ln x} \): \[ f(x) = 8x - x \].
04

Combine Like Terms

Subtract the terms to simplify the expression: \[ f(x) = 7x \].

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Key Concepts

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

Logarithmic Properties
Logarithms are a mathematical tool that help us work with very large or very small numbers by turning multiplication and division into addition and subtraction. The most common logarithms you will encounter are common logarithms (base 10) and natural logarithms (base \( e \), where \( e \approx 2.718 \)).

The natural logarithm, written as \( \ln(x) \), has several important properties that can simplify calculations. For instance:

  • Product Rule: \( \ln(ab) = \ln(a) + \ln(b) \)
  • Quotient Rule: \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \)
  • Power Rule: \( \ln(a^b) = b\ln(a) \)
  • Inverse Relationship: If \( e^y = x \), then \( \ln(x) = y \) and vice versa.
These properties are crucial for simplifying logarithmic expressions and solving equations involving exponentials. By recognizing how logarithms capture the essence of multiplication, division, and exponentiation, you can transform complex problems into simpler ones.
Exponential Functions
Exponential functions, denoted as \( e^x \), involve the constant \( e \) which is roughly 2.718. These functions are vital for modeling growth and decay processes, such as population growth, radioactive decay, and even interest rates.

One interesting characteristic of exponential functions is their rapid growth. As \( x \) increases, \( e^x \) grows very quickly compared to linear or polynomial functions. This ability to represent huge changes is what makes exponential functions so powerful in real-world applications.

When dealing with exponential functions, we often use their inverse, the natural logarithm, to simplify expressions or to solve equations. In the realm of calculus and beyond, mastering exponential functions paves the way for understanding more complex mathematical concepts.
Inverse Operations
Inverse operations are actions that "undo" each other. A great example is the inverse relationship between exponentials and logarithms. In our original exercise, we looked at \( e^{\ln x} \).

Here's how they work together:
  • The exponential function \( e^y \) grows very fast and its inverse, \( \ln(x) \), helps "bring it down" to simpler terms.
  • When you apply an exponential and its inverse to the same base, they cancel out, such as \( e^{\ln(x)} = x \). This means if you start with \( x \), use a natural log, and then an exponential, you end up where you started.
Understanding inverse operations is essential as they allow you to solve for unknowns in equations and simplify complex expressions easily. It's one of the key strategies in algebra and analysis to deal with challenging problems efficiently.

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Most popular questions from this chapter

The demand function for a newspaper is \(D(p)=80,000 \sqrt{75-p}\), where \(p\) is the price in cents. The publisher currently charges 50 cents, and it plans to raise the price to increase revenues. Will this strategy succeed?

In 1987 Carl Lewis set a new world's record of \(9.93\) seconds for the 100 -meter run. The distance that he ran in the first \(x\) seconds was $$ 11.274\left[x-1.06\left(1-e^{-x / 1.06}\right)\right] \text { meters } $$ for \(0 \leq x \leq 9.93\). Enter this function as \(y_{1}\), and define \(y_{2}\) as its derivative (using NDERIV), so that \(y_{2}\) gives his velocity after \(x\) seconds. Graph them on the window \([0,9.93]\) by \([0,100]\). Define \(y_{3}\) as the derivative of \(y_{2}\) (using NDERIV) so that \(y_{3}\) gives his acceleration after \(x\) seconds, and graph \(y_{2}\) and \(y_{3}\) on the window \([0,9.93]\) by \([0,20] .\) Evaluate both \(y_{2}\) and \(y_{3}\) at \(x=0.1\) and also at \(x=9.93\) (using EVALUATE). Interpret your answers.

For each function: a. Find the relative rate of change. b. Evaluate the relative rate of change at the given value(s) of \(t\). $$ f(t)=100 e^{-0.5 t}, \quad t=4 $$

For each function: a. Find the relative rate of change. b. Evaluate the relative rate of change at the given value(s) of \(t\). $$ f(t)=e^{-t^{2}}, \quad t=10 $$

A mug of beer chilled to 40 degrees, if left in a 70 -degree room, will warm to a temperature of \(T(t)=70-30 e^{-3.5 t}\) degrees in \(t\) hours. a. Find \(T(0.25)\) and \(T^{\prime}(0.25)\) and interpret your answers. b. Find \(T(1)\) and \(T^{\prime}(1)\) and interpret your answers.

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