/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 Use a graphing utility to solve ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use a graphing utility to solve each equation for \(x.\) $$100=50 e^{0.06 x}$$

Short Answer

Expert verified
Solving the given equation, the value of \(x\) is approximately \(11.55\).

Step by step solution

01

Understanding the equation

Given exponential equation is \(100 = 50e^{0.06x}\). This equation involves the exponential function, where the variable \(x\) is found in the exponent. Here, 'e' is Euler's number, which is approximately equal to 2.71828.
02

Isolate the Term with \(x\)

To make the calculation easier, isolate the term with \(x\) first. To do this, divide both sides of the equation by 50. This results in \(2 = e^{0.06x}\).
03

Apply the Natural Logarithm

Apply the natural logarithm \(\ln\) to both sides of the equation. Using the property \(\ln(e^a) = a\), the equation will then be \(\ln(2) = 0.06x\).
04

Solve for \(x\)

Lastly, solve for \(x\) by dividing \(\ln(2)\) by \(0.06\). This implies that \(x = \frac{\ln(2)}{0.06}\).

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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 mathematical expressions where a constant base is raised to a variable exponent. In the equation from the exercise, 100 = 50e^{0.06x}, the function f(x) = 50e^{0.06x} is an exponential function. Here, the base is Euler's number e, which is approximately 2.71828, and it is raised to the power of 0.06x.

The distinct property of exponential functions is that they change at rates proportional to their value, which is why they frequently appear in applications involving growth or decay, such as population studies or radioactive decay. In solving exponential equations, it's critical to isolate the term that contains the exponent, as done by dividing both sides of the equation by 50 in our example. This is the first step to simplifying the equation and preparing it for further operations like taking logarithms.
Natural Logarithm
The natural logarithm, denoted as ln, is the inverse operation of exponentiation when the base is e. That means if you have e^a, taking the natural logarithm of both sides, you get ln(e^a) = a. This property is pivotal when solving exponential equations because it allows us to 'undo' the exponentiation, thereby isolating the variable.

In our exercise, once we obtained the simplified equation 2 = e^{0.06x}, applying the natural logarithm to both sides gave us ln(2) = 0.06x. This equation is much simpler to handle because it transformed an exponential equation into a linear one, which can be solved for x with basic algebraic manipulation. The solution involves dividing ln(2) by 0.06 to find the value of x.
Graphing Utilities
Graphing utilities, such as graphing calculators or computer software, are tools that allow for the visualization of mathematical equations and functions. They are invaluable in solving complicated equations because they provide a visual representation that can help identify solutions.

In the context of our original exercise, a graphing utility can be used to graph the exponential function and visually find the point where it intersects with y=100, giving us the solution for x. While the analytical solution is important for understanding the process involved in solving exponential equations, graphing utilities offer a practical verification tool to ensure that our analytic solution is correct. They can also be particularly helpful when the equations are too complex to solve analytically or when a rough estimate of the solution is sufficient.

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Most popular questions from this chapter

The spread of a disease can be modcled by a logistic function. For example, in carly 2003 there was an outbreak of an illness called SARS (Severe Acute Respiratory Syndrome) in many parts of the world. The following table gives the total momber of cases in Canada for the wecks following March 20,2003 (Source: World Health Organization) (Note: The total number of cases dropped from 149 to 140 between weeks 3 and 4 because some of the cases thought to be SARS were reclassified as other discases.) $$\begin{array}{|c|c|}\hline\text { Weeks since } & \\\\\text { March } 20,2003 & \text { Total Cases } \\\0 & 9 \\\1 & 62 \\\2 & 132 \\\3 & 149 \\\4 & 140 \\\5 & 216 \\\6 & 245 \\\7 & 252 \\\8 & 250 \\\\\hline\end{array}$$ (a) Explain why a logistic function would suit this data well. (b) Make a scatter plot of the data and find the logistic function of the form \(f(x)=\frac{\epsilon}{1+a \varepsilon^{-1}}\) that best fits the data. (c) What docs \(c\) signify in your model? (d) The World Health Organization declared in July 2003 that SARS no longer posed a threat in Canada. By analyring this data, explain why that would be so.

Evaluate the expression to four decimal places using a calculator. $$\log 1400$$

This set of exercises will draw on the ideas presented in this section and your general math background. Explain why the equation \(2 e^{x}=-1\) has no solution.

Find the interest rate \(r\) if the interest on the initial deposit is compounded continuously and no withdrawals or further deposits are made. Initial amount: \(\$ 3000 ;\) Amount in 3 years: \(\$ 3600\)

The following table gives the price per barrel of crude oil for selected years from 1992 to 2006 (Source: www.ioga.com/special/crudeoil-Hist.htm) $$\begin{array}{|c|c|}\hline\text { Year } & \begin{array}{c}\text { Price } \\\\\text { (dollars) }\end{array} \\\\\hline 1992 & 19.25 \\\1996 & 20.46 \\\2000 & 27.40 \\\2004 & 37.41 \\\2006 & 58.30\\\ \hline\end{array}$$ (a) Make a scatter plot of the data and find the exponential function of the form \(P(t)=C a^{t}\) that best fits the data. Let \(t\) be the number of years since 1992 (b) Using your model, what is the projected price per barrel of crude oil in \(2009 ?\)

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