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Solve the exponential equation algebraically. Approximate the result to three decimal places.\(-14+3 e^{x}=11\)

Short Answer

Expert verified
The solution to the equation is approximately \(x = 2.120\) to three decimal places.

Step by step solution

01

Simplify the equation on both sides

First, compile like terms on both sides of the equation. Add 14 to both sides to isolate \(3e^x\) on one side: \(-14 + 14 + 3e^x = 11 + 14\), which simplifies to \(3e^x = 25\).
02

Isolate \(e^x\)

Then divide both sides of the equation by 3 to isolate \(e^x\): \(e^x = 25 / 3 = 8.333\).
03

Apply natural logarithm to both sides

Since \(e^x\) is the inverse operation of the natural logarithm (ln), apply ln to both sides of the equation to remove the exponential: \(ln(e^x) = ln(8.333)\). This simplifies to \(x = ln(8.333)\).
04

Evaluate the natural logarithm

Using a calculator or computer, compute the natural logarithm to three decimal places: \(x = ln(8.333) \approx 2.120\)

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Key Concepts

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

Algebraic Methods for Solving Exponential Equations
Understanding algebraic methods is essential for solving equations most efficiently. These methods encompass a variety of steps that strategically manipulate the equation to isolate the variable in question. When dealing with exponential equations like (-14+3e^x=11), the aim is to rewrite the equation so the exponential part, e^x, stands alone. This involves combining like terms and using arithmetic operations - addition, subtraction, multiplication, and division - to eliminate coefficients and other terms.

For instance, by adding 14 to both sides to cancel out the -14, you can first simplify the equation. After simplifying, the equation becomes more manageable: (3e^x = 25). The variable is still not isolated because it's part of the term 3e^x, prompting the next algebraic step which is division. Dividing both sides by 3 isolates e^x.
Utilizing Natural Logarithm to Solve for x
The natural logarithm, denoted as ln, serves as a crucial tool in solving exponential equations. It's the inverse operation of taking an exponential with the base e, and hence, applying it correctly simplifies equations involving e^x.

With the equation (e^x = 8.333), the use of ln enables you to revert the exponent back to the original variable. By applying ln to both sides - (ln(e^x) = ln(8.333)), we can use the identity ln(e^x) = x, which drastically simplifies the calculation. It's essential to understand that ln and e are intrinsically linked: the natural logarithm of e^x will always yield x because they are inverse functions. This property is what allows us to take the complex expression e^x and convert it into the much more manageable x.
Isolation of Variables in Equations
Isolation of the variable is the ultimate goal when solving any algebraic equation. This process involves moving all instances of the variable to one side of the equation and everything else to the opposite side.

In the exponential equation, (e^x = 8.333), our goal is to isolate x. Initially, x is part of the exponent in the term e^x, which makes it not isolated. To successfully isolate x, the equation is transformed through the application of natural logarithm as explained previously. This step effectively strips away the e, leaving x by itself on one side of the equation. Once the variable is isolated, you easily obtain its value by performing the necessary arithmetic or logarithmic calculations. Achieving this isolation is crucial because only when we have the variable alone can we determine its value and thus solve the equation.

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Most popular questions from this chapter

Population The population \(P\) of the United States officially reached 300 million at about 7:46 A.M. E.S.T. on Tuesday, October 17,2006 . The table shows the U.S. populations (in millions) since 1900. (Source: U.S. Census Bureau)$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Population } \\ \hline 1900 & 76 \\ \hline 1910 & 92 \\ \hline 1920 & 106 \\ \hline 1930 & 123 \\ \hline 1940 & 132 \\ \hline 1950 & 151 \\ \hline \end{array} $$$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Population } \\ \hline 1960 & 179 \\ \hline 1970 & 203 \\ \hline 1980 & 227 \\ \hline 1990 & 250 \\ \hline 2000 & 282 \\ \hline 2006 & 300 \\ \hline \end{array} $$(a) Use a graphing utility to create a scatter plot of the data. Let \(t\) represent the year, with \(t=0\) corresponding to 1900 . (b) Use the regression feature of a graphing utility to find an exponential model for the data. Use the Inverse Property \(b=e^{\ln b}\) to rewrite the model as an exponential model in base \(e\). (c) Graph the exponential model in base \(e\) with the scatter plot of the data. What appears to be happening to the relationship between the data points and the regression curve at \(t=100\) and \(t=106 ?\) (d) Use the regression feature of a graphing utility to find a logistic growth model for the data. Graph each model using the window settings shown below. Which model do you think will give more accurate predictions of the population well beyond \(2006 ?\)

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