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Chapter 13: Problem 45

Express \(\ln \left(e^{2} \sqrt{1-x}\right)\) as the sum or difference of logarithms, evaluating where possible.

Short Answer

Expert verified
\( 2 + \frac{1}{2}\ln(1-x) \)

Step by step solution

01

Apply Logarithm Property for Products

Begin by utilizing the property of logarithms that states \( \ln(AB) = \ln(A) + \ln(B) \) where \( A = e^2 \) and \( B = \sqrt{1-x} \). This allows us to write:\[ \ln \left(e^2 \sqrt{1-x}\right) = \ln(e^2) + \ln(\sqrt{1-x}) \]
02

Simplify the Exponentiated Term

Recognize that \( \ln(e^2) \) simplifies using the property \( \ln(e^a) = a \). Therefore:\[ \ln(e^2) = 2 \]
03

Simplify the Square Root Term

Use the property of logarithms that states \( \ln(\sqrt{x}) = \frac{1}{2} \ln(x) \). So for \( \ln(\sqrt{1-x}) \), we have:\[ \ln(\sqrt{1-x}) = \frac{1}{2} \ln(1-x) \]
04

Combine the Results

Combine the simplified terms from Steps 2 and 3 to express the original logarithm as a sum:\[ \ln \left(e^2 \sqrt{1-x}\right) = 2 + \frac{1}{2} \ln(1-x) \]

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Key Concepts

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

Product Property of Logarithms
The product property of logarithms is a fundamental rule that simplifies logarithmic expressions involving products. When you have a logarithm of a product, such as \( \ln(AB) \), it can be expressed as the sum of the logarithms of the individual factors, \( \ln(A) + \ln(B) \). This property allows for easier manipulation and simplification of logarithmic expressions.
For example, in the original exercise, the expression \( \ln(e^2 \sqrt{1-x}) \) applies this property:
  • First, identify the components of the product \( A = e^2 \) and \( B = \sqrt{1-x} \).
  • Then, apply the product property: \( \ln(e^2 \sqrt{1-x}) = \ln(e^2) + \ln(\sqrt{1-x}) \).
Breaking down the logarithm in this way simplifies further calculations and makes complex expressions more manageable.
Simplifying Exponents
Simplifying expressions involving exponents is crucial in reducing complexity, especially in logarithms. A special property of logarithms with base \( e \) (natural logarithms) simplifies exponentiated terms. For any expression \( \ln(e^a) \), it simplifies directly to \( a \). This rule comes from the natural logarithm being the inverse of the exponential function with base \( e \).
  • In our exercise, simplifying \( \ln(e^2) \) is straightforward due to this property: \( \ln(e^2) = 2 \).
This simple computation eliminates the need for further transformations, helping streamline calculations. Recognizing and applying this property can significantly expedite solving problems involving exponentiated expressions in logarithmic contexts.
Simplifying Square Roots
When dealing with square roots inside a logarithm, it's important to use the appropriate logarithmic properties to make expressions simpler. Specifically, the logarithm of a square root \( \ln(\sqrt{x}) \) can be rewritten using the power rule of logarithms as \( \frac{1}{2} \ln(x) \). This transformation considers the square root as a power \( x^{1/2} \) and then uses the power rule: \( \ln(x^a) = a \cdot \ln(x) \).
  • In the given exercise, \( \ln(\sqrt{1-x}) \) becomes \( \frac{1}{2} \ln(1-x) \).
Transforming the square root in this manner helps us handle more complex expressions by reducing them into simpler forms, making logarithmic math much more approachable and the arithmetic less daunting.

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Most popular questions from this chapter

Plot the indicated semilogarithmic graphs for the following application. In a particular electric circuit, called a low-pass filter, the input voltage \(V_{i}\) is across a resistor and a capacitor, and the output voltage \(V_{0}\) is across the capacitor (see Fig. 13.28 ). The voltage gain \(G\) (in \(d B\) ) is given by $$G=20 \log \frac{1}{\sqrt{1+(\omega T)^{2}}}$$ where \(\tan \phi=-\omega T\) Here, \(\phi\) is the phase angle of \(V_{0} / V_{i .}\) For values of \(\omega T\) of 0.01,0.1,0.3 \(1.0,3.0,10.0,30.0,\) and \(100,\) plot the indicated graphs. These graphs are called a Bode diagram for the circuit. Calculate values of \(\phi\) (as negative angles) for the given values of \(\omega T\) and plot a semilogarithmic graph of \(\phi\) vs. \(\omega T\)

\(\text {Solve the given problems.}\) When a person ingests a medication capsule, it is found that the rate \(R\) (in \(\mathrm{mg} / \mathrm{min}\) ) that it enters the bloodstream in time \(t\) (in \(\min\) ) is given by \(\log _{10} R-\log _{10} 5=t \log _{10} 0.95 .\) Solve for \(R\) as a function of \(t\)

If \(\ln x=3\) and \(\ln y=4,\) find \(\sqrt{x^{2} y}\).

Use a calculator to solve the given equations. An Earth satellite loses \(0.1 \%\) of its remaining power each week. An equation relating the power \(P\), the initial power \(P_{0}\), and the time \(t\) (in weeks) is \(\ln P=t \ln 0.999+\ln P_{0} .\) Solve for \(P\) as a function of \(t.\)

Use a calculator to solve the given equations. The curve in which a uniform wire or rope hangs under its own weight is called a catenary. An example of a catenary that we see every day is a wire strung between utility poles, as shown in Fig. 13.20. For a particular wire, the equation of the catenary it forms is \(y=2\left(e^{x / 4}+e^{-x / 4}\right),\) where \((x, y)\) is a point on the curve. Find \(x\) for \(y=5.8 \mathrm{m}.\)
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