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Given \(\ln 4=1.3863\) and \(\ln 5=1.6094\), find each of (he following. Do not use a calculator.$$ \ln \sqrt{e^{8}} $$

Short Answer

Expert verified
The value is 4.

Step by step solution

01

Simplify the expression under the logarithm

Recognize that \( \sqrt{e^{8}} = e^{8/2} = e^{4} \).
02

Apply the logarithm to simplify further

Use the property of logarithms \( \ln(e^{a}) = a \) to simplify \( \ln(e^{4}) \) to 4.
03

State the final result

Using the property from step 2, we find \( \ln \sqrt{e^{8}} = 4 \).

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Key Concepts

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

natural logarithm
To understand the solution to our exercise, it's essential to first grasp what a natural logarithm is. The natural logarithm, typically written as \(\ln\)\, is a logarithm that has the base \(\text{e}\)\. The constant \(\text{e}\)\ approximates to 2.71828 and is known as Euler's number. The natural logarithm of a number is the power to which \(\text{e}\)\ must be raised to equal that number. For instance, \(\ln(e^{x}) = x\)\.
The natural logarithm is particularly useful in mathematics since it makes certain calculations and use-cases very convenient. You'll often encounter it in calculus, complex analysis, and applied mathematics. Knowing this helps us easily understand properties and solve problems involving logarithms.
properties of logarithms
Understanding the properties of logarithms is crucial. They allow us to simplify and manipulate logarithmic expressions effectively. Here are some important properties:
  • Product Rule: \(\ln(ab) = \ln(a) + \ln(b)\)\
  • Quotient Rule: \(\ln(a/b) = \ln(a) - \ln(b)\)\
  • Power Rule: \(\ln(a^{b}) = b \ln(a)\)\
  • Inverse: \(\ln(e^{a}) = a\)\ and \(\ln(1) = 0\)\
  • Change of Base Formula: \(\ln_a(b) = \frac{\ln(b)}{\ln(a)}\)\

These properties are used to break down complex logarithmic expressions into simpler parts. In our exercise, we prominently utilize the Power Rule and the Inverse properties.
logarithmic simplification
Logarithmic simplification often involves using the properties of logarithms to make expressions more manageable. Let's consider our example from the exercise. The given problem is \(\ln \sqrt{e^{8}}\)\. To simplify this:
  • Simplify the expression inside the logarithm: Recognize \(\sqrt{e^{8}} = e^{8/2} = e^{4}\)\.
  • Use the logarithm properties: Next, apply \(\ln(e^{a}) = a\)\ to simplify \(\ln(e^{4})\) to 4.
  • State the result: Finally, \(\ln \sqrt{e^{8}} = 4\).

This shows how a daunting expression becomes simple by understanding and applying the right properties. These steps make logarithmic simplification effective and efficient.

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Most popular questions from this chapter

Differentiate. $$ y=x \sqrt[3]{x^{2}} e^{x+1 / x} $$

In an art class, students were tested at the end of the course on a final exam. Then they were retested with an equivalent test at subsequent time intervals. Their scores after time \(t\), in months, are given in the following table. $$ \begin{array}{|c|c|} \hline \text { Time, } t \text { (in months) } & \text { Score, } y \\ \hline 1 & 84.9 \% \\ 2 & 84.6 \% \\ 3 & 84.4 \% \\ 4 & 84.2 \% \\ 5 & 84.1 \% \\ 6 & 83.9 \% \\ \hline \end{array} $$ a) Use the REGRESSION feature on a grapher to fit a logarithmic function \(y=a+b \ln x\) to the data. b) Use the function to predict test scores after \(8 \mathrm{mo} ; 10 \mathrm{mo} ; 24 \mathrm{mo} ; 36 \mathrm{mo}\) c) After how long will the test scores fall below \(82 \% ?\) d) Find the rate of change of the scores and interpret its meaning.

Pharmaceutical firms spend an immense amount of money in order to test a new medication. After the drug is approved by the Federal Drug Administration, it still takes time for physicians to fully accept and make use of the medication. The use approaches a limiting value of \(100 \%\), or 1 , after time \(t\), in months. Suppose that for a new cancer medication the percentage \(P\) of physicians using the product after \(t\) months is given by $$ P(t)=100 \%\left(1-e^{-0,4 t}\right) $$ a) What percentage of doctors have accepted the medication after 0 mo? 1 mo? 2 mo? 3 mo? 5 mo? 12 mo? 16 mo? b) Find the rate of change \(P^{\prime}(t) .\) c) Sketch a graph of the function.

The initial weight of a starving animal is \(W_{0}\). Its weight \(W\) after \(t\) days is given by $$ W=W_{0} e^{-0.009 t} $$ a) What percentage of its weight does it lose each day? b) What percentage of its initial weight remains after 30 days?

Differentiate. $$ y=\frac{e^{3 t}-e^{7 t}}{e^{4 t}} $$

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