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Chapter 3: Problem 11

Solving a Simple Equation. $$\ln x=-1$$

Short Answer

Expert verified
The solution to the equation \(\ln x=-1\) is \(x=1/e\).

Step by step solution

01

Identifying the given equation

The given equation is \(\ln x=-1\). This is a logarithmic equation with base e (natural logarithm).
02

Using the inverse of the natural logarithm

The inverse of the natural logarithm (\(\ln\)) is the exponential function with base e (\(e^x\)). Therefore, we can re-arrange the equation from logarithmic to exponential form.
03

Conversion to exponential form

When we convert the equation \(\ln x=-1\) from logarithmic to exponential form, it will be \(e^{-1}=x\).
04

Simplifying the right-hand side

The number \(e\) is a mathematical constant equal to approximately 2.71828. The expression \(e^{-1}\) is in the form of a negative exponent, which means it is the reciprocal of \(e\). Thus, \(e^{-1}=1/e\).

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Key Concepts

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

Natural Logarithm
The natural logarithm, often denoted as \( \ln \), is a logarithm with a special base: the mathematical constant \( e \). The constant \( e \) is approximately equal to 2.71828. This type of logarithm is integral in many fields, including mathematics, engineering, and the sciences. The natural logarithm has unique properties that make it especially useful:
  • Base \( e \): Unlike the common logarithm (base 10), the natural logarithm uses the base \( e \).
  • Inverse Relationship: The natural logarithm is the inverse of the exponential function, which we'll delve into more in the next section.
  • Simplifies Calculations: Working with natural logarithms can sometimes turn complex multiplication and division into simpler addition and subtraction problems.
To solve a natural logarithmic equation like \( \ln x = -1 \), one would use the inverse relationship, specifying that if \( \ln x = y \), then \( e^y = x \). Thus, converting \( \ln x = -1 \) leads to \( x = e^{-1} \).
Exponential Function
The exponential function is a transcendentally important function in mathematics, symbolized as \( e^x \). This represents the function where the base is \( e \), the natural exponential constant approximately equal to 2.71828. Exponential functions are key in describing growth and decay processes in natural and social sciences.
  • Base Impact: The base \( e \) makes this function very natural and smooth. This is due to a unique property of \( e \), providing an instantaneous rate of change that is proportional to the function's current value.
  • Growth and Decay: Exponential functions model a wide range of phenomena, such as population growth, radioactive decay, and interest calculations.
  • Inverse of \( \ln \): Beyond expressing growth and decay, the exponential function serves as the inverse of the natural logarithm, allowing reconversion from logarithmic expressions back to linear form.
  • Solving Equations: If you have \( \ln x = -1 \), converting to an exponential function becomes \( e^{-1} = x \), simplifying the calculation to find \( x \).
Inverse Operations
Inverse operations are mathematical actions that reverse the effect of other operations. They are crucial for solving equations, as they can transform complicated expressions into simple ones.
  • Inverse of Logarithms: The inverse operation of taking a natural logarithm is exponentiation. Given \( \ln x = y \), applying the exponential function (base \( e \)) undoes the logarithm, turning it into \( x = e^y \).
  • Simplifying Equations: Recognizing inverse operations helps one solve equations more quickly. For example, with \( \ln x = -1 \), multiplication by \( e \) simplifies the calculation to direct evaluation: \( x = e^{-1} \).
  • Broader Applications: This concept applies to other operations as well, such as addition and subtraction or division and multiplication. Understanding how operations interact in reverse enables problem-solving across various mathematical disciplines.
Applying inverse operations can drastically simplify solving equations. In our equation \( \ln x = -1 \), using the inverse property, we apply \( e \) to both sides to find \( x = e^{-1} \). This results in a straightforward calculation to determine \( x \).

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Most popular questions from this chapter

Find a logarithmic equation that relates \(y\) and \(x .\) Explain the steps used to find the equation. $$\begin{array}{|c|c|c|c|c|c|c|}\hline x & 1 & 2 & 3 & 4 & 5 & 6 \\\\\hline y & 1 & 1.587 & 2.080 & 2.520 & 2.924 & 3.302 \\ \hline\end{array}$$

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

Rewrite each verbal statement as an equation. Then decide whether the statement is true or false. Justify your answer. The logarithm of the product of two numbers is equal to me sum of the logarithms of the numbers.

Use the change-of-base formula to rewrite the logarithm as a ratio of logarithms. Then use a graphing utility to graph the ratio. $$f(x)=\log _{2} x$$

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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