/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 Simplify the expression. $$ ... [FREE SOLUTION] | 91Ó°ÊÓ

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Simplify the expression. $$ \ln \left(\frac{1}{e^{3}}\right) $$

Short Answer

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Step by step solution

01

Understand the Expression

Write down the given expression: \ \ \(\text{ln} \bigg(\frac{1}{e^{3}} \bigg) \)
02

Apply the Property of Logarithms

Use the property that \(\text{ln} \bigg(\frac{1}{x} \bigg) = -\text{ln}(x)\). In this case, \(x = e^{3}\). Therefore, \ \(\text{ln} \bigg(\frac{1}{e^{3}} \bigg) = -\text{ln}(e^{3})\)
03

Simplify the Logarithm

Recall the property of logarithms that \(\text{ln}(e^{x}) = x\). Use this property here where \(x = 3\): \ \(-\text{ln}(e^{3}) = -3\)
04

Write the Final Answer

Combining all the simplifications, the expression simplifies to: \ \(\text{ln} \bigg(\frac{1}{e^{3}} \bigg) = -3\)

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Key Concepts

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

Logarithmic Properties
Understanding logarithmic properties makes handling logarithms much easier. One fundamental property is that the natural logarithm (ln) of a fraction is the negative of the natural logarithm of its reciprocal. For example, \(\text{ln} \bigg(\frac{1}{x}\bigg) = -\text{ln}(x)\). This is a crucial property that helps in simplifying logs of fractions.
Another important property is related to the base of the natural logarithms, 'e'. The natural logarithm of \(e^x\) simplifies to \(x\) itself. So, \(\text{ln}(e^x) = x\). Knowing these properties allows us to break down complex logarithmic expressions into simpler parts that are easier to solve. Let's dive into how these properties apply in real-world problems!
Logarithm Simplification
Simplifying logarithmic expressions involves using known properties to break them down. Let's take an example: Consider \(\text{ln} \bigg(\frac{1}{e^3}\bigg)\). Start by applying the property \(\text{ln} \bigg(\frac{1}{x}\bigg) = -\text{ln}(x)\). Hence, \(\text{ln} \bigg(\frac{1}{e^3}\bigg)\) will simplify to \(-\text{ln}(e^3)\).
Next, use the property \(\text{ln}(e^x) = x\). Here, \(x\) is 3, so \(-\text{ln}(e^3) = -3\).
The properties allow us to simplify the expression to \(\text{ln} \bigg(\frac{1}{e^3}\bigg) = -3\). These steps show how useful logarithmic properties can be for simplification.
Exponential Functions
Exponential functions are critical in many areas of math and science. They are of the form \(e^x\), where 'e' is Euler's number, approximately equal to 2.71828.
When working with natural logarithms, understanding exponential functions becomes necessary because natural logs are based on the exponential function 'e'.
For example, the expression \(\text{ln}(e^x) = x\) stems from the unique relationship where the natural logarithm and the exponential function cancel each other out. So, if you ever see \(\text{ln}(e^x)\), you can directly simplify it to \(x\).
This helps solve various problems involving growth and decay models, like population growth or radioactive decay, where exponential functions naturally arise.

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Most popular questions from this chapter

A table of data is given. a. Graph the points and from visual inspection, select the model that would best fit the data. Choose from $$\begin{array}{ll} y=m x+b \text { (linear) } & y=a b^{x} \text { (exponential) } \\ y=a+b \ln x \text { (logarithmic) } & y=\frac{c}{1+a e^{-b x}} \text { (logistic) } \end{array}$$ b. Use a graphing utility to find a function that fits the data. $$ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 2.3 \\ \hline 4 & 3.6 \\ \hline 8 & 5.7 \\ \hline 12 & 9.1 \\ \hline 16 & 14 \\ \hline 20 & 22 \\ \hline \end{array} $$

Two million \(E\). coli bacteria are present in a laboratory culture. An antibacterial agent is introduced and the population of bacteria \(P(t)\) decreases by half every \(6 \mathrm{hr}\). The population can be represented by \(P(t)=2,000,000\left(\frac{1}{2}\right)^{t / 6}\) a. Convert this to an exponential function using base \(e\). b. Verify that the original function and the result from part (a) yield the same result for \(P(0), P(6), P(12)\), and \(P(60) .\) (Note: There may be round- off error.)

The population of Canada \(P(t)\) (in millions) since January \(1,1900,\) can be approximated by $$P(t)=\frac{55.1}{1+9.6 e^{-0.02515 t}}$$ where \(t\) is the number of years since January 1,1900 . a. Evaluate \(P(0)\) and interpret its meaning in the context of this problem. b. Use the function to predict the Canadian population on January \(1,2015 .\) Round to the nearest million. c. Use the function to predict the Canadian population on January 1,2040 . d. Determine the year during which the Canadian population will reach 45 million. e. What value will the term \(\frac{9.6}{e^{0.02515 t}}\) approach as \(t \rightarrow \infty\) ? f. Determine the limiting value of \(P(t)\).

Write \(10^{2 x-4}=80,600\) in logarithmic form.

Solve the equation. Write the solution set with the exact values given in terms of common or natural logarithms. Also give approximate solutions to 4 decimal places. \(1024=19^{x}+4\)

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