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Chapter 3: Problem 62

Evaluate \(g(x)=\ln x\) at the indicated value of \(x\) without using a calculator. $$x=e^{-4}$$

Short Answer

Expert verified
-4

Step by step solution

01

Substitute the value of x

We start off by substituting the value of \(x\) in \(g(x)\). Thus, we have \[g(x)=\ln(e^{-4})\].
02

Use the property of logarithm

The base of natural logarithm is \(e\). And the logarithm base \(b\) of a number \(n\) that equals \(b\) raised to the power \(n\) is equal to \(n\). Thus, when we take natural logarithm of \(e^{a}\), it equals to \(a\). Therefore, \(\ln(e^{-4}) = -4\).

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Key Concepts

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

Logarithmic Properties
Logarithms have several fascinating properties that simplify complex calculations, especially when dealing with exponential expressions.
The natural logarithm, denoted as \( \ln \), is a special logarithm where the base is the mathematical constant \( e \), approximately equal to 2.718. This is a crucial concept to recognize.
  • Logarithm of a Power: For any number \( a \), the natural logarithm of its power, \( \ln(e^a) \), is equal to \( a \). This is because exponential and logarithmic functions are inverses of each other.
  • Property of Inverses: Consider \( \ln(b^c) = c \ln(b) \). This property allows us to simplify expressions where a logarithm is applied to an exponent.
  • Logarithm of a Product: When multiplying numbers, the logarithm turns it into an addition: \( \ln(xy) = \ln(x) + \ln(y) \).
  • Logarithm of a Quotient: Dividing turns into subtraction: \( \ln(\frac{x}{y}) = \ln(x) - \ln(y) \).
Understanding and applying these properties can greatly ease the process of solving logarithmic and exponential equations efficiently.
Exponential Functions
Exponential functions are an essential topic in mathematics and have a unique way of growing or decaying. They are expressed in the form \( f(x) = b^x \), where \( b \) is the base, and \( x \) is the exponent.
For the function \( g(x) = e^x \), the base is \( e \). This function is widely used in modeling real-world phenomena such as population growth, radioactive decay, and interest calculations.
  • Growth and Decay: If \( b > 1 \), \( f(x) \) represents exponential growth. If \( 0 < b < 1 \), it represents exponential decay.
  • Natural Exponential Functions: With \( e \) as the base, these functions are key in calculus, particularly in finding derivatives and integrals.
  • Inverse Nature with Logarithms: Exponential functions are the inverses of logarithmic functions. This means that \( e^{\ln(x)} = x \).
Grasping the behavior and characteristics of exponential functions can significantly aid in solving problems involving these expressions.
Evaluation of Logarithms
Evaluating logarithms often involves understanding and utilizing the properties of logarithms. Let's break down how to evaluate a logarithm like \( \ln(x) \).
Consider the specific case of evaluating \( \ln(e^{-4}) \).
  • Substitute the Expression: Identify the argument of the logarithm. In this case, we substitute \( e^{-4} \) into \( \ln \), giving \( \ln(e^{-4}) \).
  • Apply the Logarithm Power Property: The property \( \ln(e^a) = a \) tells us directly that \( \ln(e^{-4}) = -4 \). This is because the logarithm to the base \( e \) of any \( e^x \) is simply \( x \).
  • Understand the Result: The simplification of \( \ln(e^{-4}) = -4 \) illustrates the inverse relationship between exponential and logarithmic functions, confirming that we've correctly evaluated the expression.
Mastering these steps in evaluating logarithms enables you to confidently handle more complex expressions and equations without needing a calculator.

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Most popular questions from this chapter

Solve the equation algebraically. Round your result to three decimal places. Verify your answer using a graphing utility. $$2 x \ln x+x=0$$

Condense the expression to the logarithm of a single quantity. $$2[3 \ln x-\ln (x+1)-\ln (x-1)]$$

Solve the equation algebraically. Round your result to three decimal places. Verify your answer using a graphing utility. $$2 x^{2} e^{2 x}+2 x e^{2 x}=0$$

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$\ln (x+5)=\ln (x-1)-\ln (x+1)$$

A cup of water at an initial temperature of \(78^{\circ} \mathrm{C}\) is placed in a room at a constant temperature of \(21^{\circ} \mathrm{C}\). The temperature of the water is measured every 5 minutes during a half-hour period. The results are recorded as ordered pairs of the form \((t, T),\) where \(t\) is the time (in minutes) and \(T\) is the temperature (in degrees Celsius). $$\begin{aligned} &\left(0,78.0^{\circ}\right),\left(5,66.0^{\circ}\right),\left(10,57.5^{\circ}\right),\left(15,51.2^{\circ}\right)\\\ &\left(20,46.3^{\circ}\right),\left(25,42.4^{\circ}\right),\left(30,39.6^{\circ}\right) \end{aligned}$$ (a) The graph of the model for the data should be asymptotic with the graph of the temperature of the room. Subtract the room temperature from each of the temperatures in the ordered pairs. Use a graphing utility to plot the data points \((t, T)\) and \((t, T-21)\). (b) An exponential model for the data \((t, T-21)\) is given by \(T-21=54.4(0.964)^{t} .\) Solve for \(T\) and graph the model. Compare the result with the plot of the original data. (c) Take the natural logarithms of the revised temperatures. Use the graphing utility to plot the points \((t, \ln (T-21))\) and observe that the points appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. This resulting line has the form \(\ln (T-21)=a t+b\) Solve for \(T,\) and verify that the result is equivalent to the model in part (b). (d) Fit a rational model to the data. Take the reciprocals of the \(y\) -coordinates of the revised data points to generate the points $$\left(t, \frac{1}{T-21}\right)$$. Use the graphing utility to graph these points and observe that they appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. The resulting line has the form $$\frac{1}{T-21}=a t+b$$. Solve for \(T,\) and use the graphing utility to graph the rational function and the original data points. (e) Why did taking the logarithms of the temperatures lead to a linear scatter plot? Why did taking the reciprocals of the temperatures lead to a linear scatter plot?
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