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Chapter 4: Problem 88

Find the real solutions to the equation. $$x^{2} e^{x}-e^{x}=0$$

Short Answer

Expert verified
The real solutions are \( x = 1 \) and \( x = -1 \).

Step by step solution

01

Factor out the common term

Notice that both terms on the left side of the equation \(x^{2} e^{x}-e^{x}=0\) have a common factor of \(e^{x}\). Factor \(e^{x}\) out of the equation.\(e^{x}(x^{2} - 1) = 0\)
02

Set each factor to zero

Since the product of two terms is zero, set each factor equal to zero separately. First, set \(e^{x} = 0\). Second, set \(x^{2} - 1 = 0\).
03

Solve \(e^{x} = 0\)

Recall that the exponential function \(e^{x}\) is never zero for any real value of \(x\). Therefore, there are no real solutions from this equation.No real solutions from \(e^{x} = 0\).
04

Solve \(x^{2} - 1 = 0\)

Add 1 to both sides of the equation to get \(x^{2} = 1\). Then take the square root of both sides to find that \( x = 1\) or \(x = -1\).
05

Combine the solutions

The real solutions to the original equation are the solutions from \(x^{2} - 1 = 0\). Therefore, the real solutions to the equation \(x^{2} e^{x}-e^{x}=0\) are \( x = 1\) or \(x = -1\).

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Key Concepts

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

factoring
When solving algebraic equations, factoring is a crucial technique. Factoring involves breaking down a complex expression into simpler components, known as factors, that when multiplied together produce the original expression.
In the equation given, \(x^{2} e^{x} - e^{x} = 0\), the common term \(e^{x}\) was factored out, transforming the equation into \(e^{x}(x^{2} - 1) = 0\).
This makes it easier to solve, as each factor can be set to zero individually:
  • \(e^{x} = 0\)
  • \(x^{2} - 1 = 0\)
Factoring simplifies the equation-solving process, especially when dealing with polynomials or various expressions in algebra.
exponential functions
Exponential functions are a type of mathematical function where a constant base, such as \(e\) (Euler's number approximately equal to 2.71828), is raised to a variable exponent. The general form is \(f(x) = a e^{bx}\), where \(a\) and \(b\) are constants.
In our exercise, the exponential function is \(e^{x}\). One key property of \(e^{x}\) is that it is never zero for any real value of \(x\).
This means when solving \(e^{x} = 0\), we conclude there are no real solutions for this part of the equation. Understanding these properties helps in solving and simplifying exponential equations in algebra.
Remember:
  • Exponential functions grow rapidly as the variable increases
  • They never equal zero for any real number
  • They have important applications in various fields like biology, finance, and physics
quadratic equations
Quadratic equations are a key topic in algebra, expressed in the form \(ax^2 + bx + c = 0\). A quadratic equation is a polynomial equation of degree 2. To solve quadratic equations, there are various methods:
In our example, the reduced part of the equation is \(x^2 - 1 = 0\), which is a quadratic equation. We solve it by isolating the variable \(x\):
  • Add 1 to both sides: \(x^2 = 1\)
  • Take the square root: \(x = 1\) or \(x = -1\)
Thus, the solutions to the quadratic part of the equation are \(1\) and \(-1\). These concepts can be applied to various problems involving quadratic equations.
Key points about quadratic equations:
  • They appear in diverse real-world scenarios such as projectile motion, area problems, and optimization tasks
  • They can have two real solutions, one real solution, or no real solutions depending on their discriminant
  • Mastery of quadratic equations is essential for progressing in algebra and higher-level math

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Most popular questions from this chapter

Solve the equation. Write the solution set with the exact solutions. Also give approximate solutions to 4 decimal places if necessary. \(\ln x+\ln (x-4)=\ln (3 x-10)\)

A table of data is given. a. Graph the points and from visual inspection, select the model that would best fit the data. Choose from $$\begin{array}{ll} y=m x+b \text { (linear) } & y=a b^{x} \text { (exponential) } \\ y=a+b \ln x \text { (logarithmic) } & y=\frac{c}{1+a e^{-b x}} \text { (logistic) } \end{array}$$ b. Use a graphing utility to find a function that fits the data. $$ \begin{array}{|c|c|} \hline x & y \\ \hline 3 & 2.7 \\ \hline 7 & 12.2 \\ \hline 13 & 25.7 \\ \hline 15 & 30 \\ \hline 17 & 34 \\ \hline 21 & 44.4 \\ \hline \end{array} $$

Use the model \(A=P e^{r t} .\) The variable \(A\) represents the future value of \(P\) dollars invested at an interest rate \(r\) compounded continuously for \(t\) years. If a couple has \(\$ 80,000\) in a retirement account, how long will it take the money to grow to \(\$ 1,000,000\) if it grows by \(6 \%\) compounded continuously? Round to the nearest year.

Determine if the given value of \(x\) is a solution to the logarithmic equation. \(\log _{4} x=3-\log _{4}(x-63)\) a. \(x=64\) b. \(x=-1\) c. \(x=32\)

(See Example 8 ) a. Estimate the value of the logarithm between two consecutive integers. For example, \(\log _{2} 7\) is between 2 and 3 because \(2^{2}<7<2^{3}\). b. Use the change-of-base formula and a calculator to approximate the logarithm to 4 decimal places. c. Check the result by using the related exponential form. $$ \log _{2} 0.2 $$
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