Problem 83 The given equations are quadrati... [FREE SOLUTION]

91影视

A Graphical Approach to Precalculus with Limits

John Hornsby, Margaret L. Lial, Gary Rockswold

$Math Studyset 91影视 Explanations$ Math

7 Edition

Chapter 6: Problem 83

The given equations are quadratic in form. Solve each and give exact solutions. $$2 e^{2 x}+e^{x}=6$$

Short Answer

Expert verified

The solution is $x = \ln\left(\frac{3}{2}\right)$.

Step by step solution

Observe the Equation

The given equation is $2e^{2x} + e^{x} = 6$. Notice that this resembles a quadratic equation in terms of $e^x$. We aim to rewrite it in that form.

Make a Substitution

Let $u = e^x$. The equation $2e^{2x} + e^{x} = 6$ can be rewritten as $2u^2 + u = 6$.

Rearrange into Quadratic Form

Rearrange the equation in standard quadratic form: $2u^2 + u - 6 = 0$.

Solve the Quadratic Equation

Now solve the quadratic equation $2u^2 + u - 6 = 0$ using either factoring, the quadratic formula, or any other method.

Use the Quadratic Formula

Apply the quadratic formula $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=2$, $b=1$, and $c=-6$. Calculate the discriminant: $b^2 - 4ac = 1^2 - 4(2)(-6) = 49$.

Calculate the Roots

Plug the values into the quadratic formula to find the roots: \[u = \frac{-1 \pm \sqrt{49}}{4}\] \[u = \frac{-1 \pm 7}{4}\] Thus, the roots are $u = \frac{6}{4} = \frac{3}{2}$ and $u = \frac{-8}{4} = -2$.

Substitute Back to Solve for x

Recall $u = e^x$. Solve $e^x = \frac{3}{2}$ and $e^x = -2$.$e^x = -2$ has no valid solution as the exponential function is always positive.

Solve for x from Valid Equation

For $e^x = \frac{3}{2}$, take the natural logarithm to solve for $x$. $x = \ln\left(\frac{3}{2}\right)$.

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Key Concepts

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

Exponential Functions

An exponential function is a mathematical expression where a constant base is raised to a variable exponent. In the problem, the exponential function is expressed as $ e^{x} $, where $ e $ is Euler鈥檚 number, approximately 2.71828. Exponential functions have unique properties:

The base $ e $ is a constant, leading to continuous growth or decay dynamics.
They grow rapidly due to the variable exponent, which changes the base's magnitude.
They are always positive. This is important because it helps determine the domain and range of the problem.

In this exercise, recognizing that $ 2e^{2x} + e^x = 6 $ is in terms of $ e^x $ allows us to utilize quadratic methods. The substitution method simplifies this equation, making it easier to solve.

Substitution Method

The substitution method is a technique used to transform complex equations into simpler forms. In this problem, the substitution $ u = e^x $ transforms the original equation into $ 2u^2 + u = 6 $. Here鈥檚 how substitution simplifies solving:

By letting $ u = e^x $, we turn an exponential equation into a quadratic one, $ 2u^2 + u - 6 = 0 $, which is easier to handle.
The main goal of substitution is to simplify the equation into a form where established methods, like the quadratic formula, can be applied.
After solving for $ u $, we revert the substitution to find the original variable, $ x $.

This method is especially helpful in transforming complex functions into a familiar algebraic format, facilitating straightforward solutions to otherwise complicated expressions.

Quadratic Formula

The quadratic formula provides an exact way to solve quadratic equations. It is defined as $ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $. This method is valuable because:

It guarantees a solution when the equation is in standard quadratic form $ au^2 + bu + c = 0 $.
The formula encompasses all potential cases for the roots (real and complex numbers).
It directly utilizes coefficients $ a $, $ b $, and $ c $, making it applicable when factoring is difficult or impossible.

In this exercise, using the quadratic formula to solve $ 2u^2 + u - 6 = 0 $, we find $ u = \frac{-1 \pm 7}{4} $, which results in roots $ u = \frac{3}{2} $ and $ u = -2 $. By knowing these roots, we can advance in finding solutions in the original problem context.

Discriminant

The discriminant is a part of the quadratic formula, $ b^2 - 4ac $, and it reveals crucial information about a quadratic equation's roots.

If the discriminant is positive, the quadratic equation has two distinct real roots.
If it's zero, there is exactly one real root, meaning the graph of the equation touches the x-axis at one point.
If negative, there are no real roots, and the solutions to the equation will be complex numbers.

In this exercise, calculating the discriminant: $ 1^2 - 4(2)(-6) = 49 $ indicates two real roots. This positive result supports the applicability of the quadratic formula in obtaining $ u = \frac{3}{2} $ and $ u = -2 $. The discriminant helps verify the nature of solutions before detailed calculations, offering clarity on expected outcomes.

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