/*! 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 17 Solve each exponential equation.... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve each exponential equation. Express irrational solutions as decimals correct to the nearest thousandth. $$e^{3 x-7} \cdot e^{-2 x}=4 e$$

Short Answer

Expert verified
x ≈ 9.386

Step by step solution

01

Combine the Exponential Terms

Use the property of exponents, which states that when multiplying like bases, you add the exponents. Combine the terms on the left side: e^{(3x-7) + (-2x)} = 4e
02

Simplify the Exponents

Simplify the exponent on the left-hand side: e^{3x-7-2x} = e^{x-7}The equation now is: e^{x-7} = 4e
03

Isolate the Exponential Term

To isolate the exponential term, notice that both sides have the constant e. We can divide both sides by e to get: e^{x-7} / e = 4 This simplifies to: e^{x-8} = 4
04

Apply the Natural Logarithm

Apply the natural logarithm ( ln )on both sides to solve for x: ln(e^{x-8}) = ln(4)By the property of logarithms, ln(e^{x-8}) = x-8. The equation is now: x-8 = ln(4)
05

Solve for x

Solve for x by adding 8 to both sides: x = ln(4) + 8 Calculate the value using a calculator: ln(4) ≈ 1.386Thus: x ≈ 1.386 + 8 ≈ 9.386

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Key Concepts

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

Exponential Equations
An exponential equation is an equation in which a variable appears in the exponent. These types of equations can sometimes be tricky, but with the right approach, they can be simplified and solved step by step. For example, in the equation given: \( e^{3x-7} \times e^{-2x} = 4e \), we start by recognizing that both terms on the left side involve the exponential base e.

Exponential equations often require skills like combining exponents, isolating terms, and using logarithms.

Understanding these steps will enhance your problem-solving skills and make exponential equations much less intimidating!
Natural Logarithm
The natural logarithm (ln) is a special type of logarithm with the base e, where e is approximately 2.71828.

Using natural logarithms is very common when solving exponential equations because the natural logarithm of e raised to a power simplifies neatly. This property is used in our example equation: \( e^{x-8} = 4 \).

To solve for x, we take the natural logarithm of both sides. This gives us:

\[ \text{ln} \big(e^{x-8}\big) = \text{ln}(4) \ \text{Since } \text{ln}(e^{x-8}) = x - 8 \ x - 8 = \text{ln}(4) \]

This transformation gets rid of the exponent, leaving a linear equation that's easier to solve. Values like \( \text{ln}(4) \) can be calculated using a calculator to get a decimal approximation, making further steps more straightforward.
Properties of Exponents
Understanding the properties of exponents is crucial for simplifying and solving exponential equations. One key property is that when you multiply like bases, you can add the exponents: \( a^m \times a^n = a^{m+n} \).

In our example, we have: \( e^{3x-7} \times e^{-2x} \). Using the property of exponents, we combine these into:

\[ e^{(3x-7) + (-2x)} = 4e \]

This helps simplify the problem, making it easier to isolate the term involving the exponent and solve for the variable. Always remember to look for opportunities to apply these properties to combine or simplify exponential terms in an equation.
Simplification
Simplification plays a key role in solving exponential equations. By combining like terms and reducing the equation to a simpler form, finding the solution becomes much more manageable.

For example, after combining the exponents in \( e^{3x-7} \times e^{-2x} \), we simplify it to: \( e^{3x-7-2x} \), which further reduces to \( e^{x-7} \).

Next, we divide both sides by e to isolate the exponential term: \( e^{x-7}/e = 4 \), which simplifies to \( e^{x-8} = 4 \). This step-by-step simplification helps isolate the variable.

Finally, applying the natural logarithm simplifies the equation from an exponential form to a linear form, making it straightforward to solve for the variable. Simplification helps break down complex steps into more manageable ones, enabling an easier path to the solution.

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Most popular questions from this chapter

Use the definition of inverses to determine whether \(f\) and \(g\) are inverses. $$f(x)=\sqrt{x+8}, \quad x \geq-8 ; \quad g(x)=x^{2}-8, \quad x \geq 0$$

For each function as defined that is one-to-one, (a) write an equation for the inverse function in the form \(y=f^{-1}(x),\) (b) graph \(f\) and \(f^{-1}\) on the same axes, and \((c)\) give the domain and the range of \(f\) and \(f^{-1}\). If the function is not one-to-one, say so. $$y=3 x-4$$

For each function as defined that is one-to-one, (a) write an equation for the inverse function in the form \(y=f^{-1}(x),\) (b) graph \(f\) and \(f^{-1}\) on the same axes, and \((c)\) give the domain and the range of \(f\) and \(f^{-1}\). If the function is not one-to-one, say so. $$f(x)=\frac{2 x+6}{x-3}, \quad x \neq 3$$

(Modeling) Solve each problem. See Example 11 . Atmospheric Pressure The atmospheric pressure (in millibars) at a given altitude (in meters) is shown in the table. $$\begin{array}{c|c||c|c} \hline \text { Altitude } & \text { Pressure } & \text { Altitude } & \text { Pressure } \\ \hline 0 & 1013 & 6000 & 472 \\ \hline 1000 & 899 & 7000 & 411 \\ \hline 2000 & 795 & 8000 & 357 \\ \hline 3000 & 701 & 9000 & 308 \\ \hline 4000 & 617 & 10,000 & 265 \\ \hline 5000 & 541 & & \\ \hline \end{array}$$ (a) Use a graphing calculator to make a scatter diagram of the data for atmospheric pressure \(P\) at altitude \(x\). (b) Would a linear or an exponential function fit the data better? (c) The function $$ P(x)=1013 e^{-0.0001341 x} $$ approximates the data. Use a graphing calculator to graph \(P\) and the data on the same coordinate axes. (d) Use \(P\) to predict the pressures at \(1500 \mathrm{m}\) and \(11,000 \mathrm{m},\) and compare them to the actual values of 846 millibars and 227 millibars, respectively.

The table shows the cost of a year's tuition, room and board, and fees at 4 -year public colleges for the years \(2002-2010 .\) Letting \(y\) represent the cost and \(x\) the number of years since 2002 we find that the function $$f(x)=9318(1.06)^{x}$$ models the data quite well. According to this function, when will the cost in 2002 be doubled? $$\begin{array}{c|c}\text { Year } & \text { Average Annual cost } \\\\\hline 2002 & \$ 9,119 \\ \hline 2003 & \$ 9,951 \\\\\hline 2004 & \$ 10,648 \\\\\hline 2005 & \$ 11,322 \\\ \hline 2006 & \$ 11,929 \\\\\hline 2007 & \$ 12,698 \\\\\hline 2008 & \$ 13,467 \\ \hline 2009 & \$ 14,265 \\\\\hline 2010 & \$ 15,067 \\\\\hline\end{array}$$

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