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91Ó°ÊÓ

Evaluate each expression. Do not use a calculator. $$\ln e^{0.5}$$

Short Answer

Expert verified
\( ln e^{0.5} = 0.5\).

Step by step solution

01

Understanding the natural logarithm

The natural logarithm function, written as \(\ln\), is the inverse of the exponential function with base \(e\). This means that \(\ln(e^x) = x\) for any real number \(x\).
02

Identify the exponent

In the given expression, \(e^{0.5}\), the exponent is \(0.5\). This is the value of \(x\) in the expression \(e^x\).
03

Apply the property of natural logarithm

Using the property \(\ln(e^x) = x\), substitute \(x\) with \(0.5\). Therefore, \(ln(e^{0.5}) = 0.5\).
04

Conclusion

By applying the property of the natural logarithm, we find that the value of the expression \(ln e^{0.5}\) is \(0.5\).

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Key Concepts

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

Exponential Functions
Exponential functions are a fundamental concept in mathematics. They are widely used in various fields, from science to finance. The general form of an exponential function is expressed as \(f(x) = a^x\), where "a" is a constant, and "x" is the exponent or power. A unique characteristic of exponential functions is that they can grow very rapidly or decay very slowly, depending on the base "a".
Most commonly, the base used is the mathematical constant \(e\), approximately equal to 2.718. When the base is \(e\), the function is referred to as the natural exponential function, denoted by \(e^x\).
Exponential growth or decay is often observed in real-world scenarios, such as population growth, radioactive decay, or compound interest. These functions model situations where a quantity increases or decreases at a rate proportional to its current value. Understanding the behavior of exponential functions can provide valuable insights into predicting and interpreting these scenarios.
Exponential functions with base \(e\) have special properties that make them easier to work with, especially when dealing with natural logarithms, which are their inverses.
Inverse Functions
Inverse functions are pairs of functions that "undo" each other. If you have a function \(f(x)\), its inverse, denoted as \(f^{-1}(x)\), is such that applying \(f^{-1}\) to \(f(x)\) will return the original value \(x\). This means \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\).
The natural logarithm, written as \(\ln(x)\), is the inverse function of the natural exponential function \(e^x\). This relationship is central to unraveling expressions that involve exponential functions, such as converting exponential growth to a linear rate of increase.
When evaluating \(\ln(e^x)\), by definition of the inverse, we derive that \(\ln(e^x) = x\). This simplification helps in complex equations involving natural logarithms and exponentials, making them much more manageable.
Understanding inverse functions is crucial when solving equations or modeling real-life situations that involve reversing a process or finding the original input from an output. They provide a critical tool for tracing back through computational processes carried out by functions.
Mathematical Properties
Mathematical properties are rules that define how numbers and operations interact. These properties serve as foundational tools for solving complex problems and simplifying expressions. In the context of exponential and logarithmic functions, there are specific properties worth highlighting.
  • Inverses and Equalities: The most vital property is that \(\ln(e^x) = x\). This property stems from the inverse relationship between the natural logarithm and the exponential function.
  • Logarithmic Identities: Another useful property is \(\ln(ab) = \ln(a) + \ln(b)\), which allows for the breakdown of logarithmic expressions into simpler parts.
  • Identity Element: Any number raised to the power of zero equals one, \(a^0 = 1\). Consequently, \(\ln(e^0) = 0\).
These properties facilitate the manipulation and understanding of more complex mathematical problems. They are essential tools for students, particularly when moving into calculus or higher-level math where exponential and logarithmic functions become more prevalent. Recognizing and applying these properties can greatly simplify tasks involving logarithms and exponentials.

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Most popular questions from this chapter

Suppose that a sample of bacteria has a concentration of 2 million bacteria per milliliter and it doubles in concentration every 12 hours. Then the time \(T\) it takes for the sample to reach a concentration of \(C\) can be approximated by the following logarithmic function. \(T(C)=\frac{500}{29} \ln \frac{C}{2}\) (a) Find the domain of \(T .\) Interpret your answer. (b) How long does it take for the concentration of bacteria to increase by \(50 \% ?\) (c) Determine the concentration \(C\) after 15 hours by solving the equation \(T(C)=15\)

The growth of bacteria in food products makes it necessary to date some products (such as milk) so that they will be sold and consumed before the bacterial count becomes too high. Suppose that, under certain storage conditions, the number of bacteria present in a product is $$f(t)=500 e^{0.1 t}$$ where \(t\) is time in days after packing of the product and the value of \(f(t)\) is in millions. (a) If the product cannot be safely eaten after the bacterial count reaches \(3,000,000,000,\) how long will this take? (b) If \(t=0\) corresponds to January \(1,\) what date should be placed on the product?

Escherichia coli is a strain of bacteria that occurs naturally in many organisms. Under certain conditions, the number of bacteria present in a colony is approximated by $$A(t)=A_{0} e^{0.023 t}$$ where \(t\) is in minutes. If \(A_{0}=2,400,000,\) find the number of bacteria at each time. Round to the nearest hundred thousand. (a) 5 minutes (b) 10 minutes (c) 60 minutes

Suppose that \(\$ 2500\) is invested in an account that pays interest compounded continuously. Find the amount of time that it would take for the account to grow to the given amount at the given rate of interest. Round to the nearest tenth of a year. \(\$ 5000\) at \(6 \%\)

Emissions Governments could reduce carbon emissions by placing a tax on fossil fuels. The costbenefit equation $$\ln (1-P)=-0.0034-0.0053 x$$ estimates the relationship between a tax of \(x\) dollars per ton of carbon and the percent \(P\) reduction in emissions of carbon, where \(P\) is in decimal form. Determine \(P\) when \(x=60 .\) Interpret the result. (Source: Clime, W., The Economics of Global Warming, Institute for International Economics.)

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