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Chapter 5: Problem 9

Find all the real-number roots of each equation. In each case, give an exact expression for the root and also (where appropriate) a calculator approximation rounded to three decimal places. $$10^{2 x}+3\left(10^{x}\right)-10=0$$

Short Answer

Expert verified
The real root is \( x = \log_{10} 2 \approx 0.301 \).

Step by step solution

01

Substitute the Variable

Let's start by making a substitution to simplify the equation. Set \( y = 10^x \). This implies that \( y^2 = 10^{2x} \). Thus, the equation becomes: \[ y^2 + 3y - 10 = 0 \].
02

Apply the Quadratic Formula

The equation \( y^2 + 3y - 10 = 0 \) is a quadratic equation in the standard form \( ay^2 + by + c = 0 \), where \( a = 1 \), \( b = 3 \), and \( c = -10 \). Apply the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].
03

Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac \): \( 3^2 - 4 \cdot 1 \cdot (-10) = 9 + 40 = 49 \). Since the discriminant is positive, there are two real solutions.
04

Find the Roots of the Quadratic

Substitute the discriminant back into the quadratic formula: \[ y = \frac{-3 \pm \sqrt{49}}{2 \times 1} = \frac{-3 \pm 7}{2} \]. Solve for \( y \):- \( y = \frac{4}{2} = 2 \)- \( y = \frac{-10}{2} = -5 \).
05

Discard the Negative Solution

Since \( y = 10^x \) and exponential functions cannot be negative, discard \( y = -5 \). We only consider \( y = 2 \).
06

Solve for x

Now that we have \( y = 2 \), solve for \( x \) by reversing the substitution: \( 10^x = 2 \). Take the logarithm of both sides to find \( x \): \( x = \log_{10} 2 \). Calculate this using a calculator.
07

Calculate the Approximate Solution

Use a calculator to find \( \log_{10} 2 \). You will find that \( \log_{10} 2 \approx 0.301 \).
08

Summarize the Solutions

Thus, the equation \( 10^{2x} + 3(10^x) - 10 = 0 \) has one real root at \( x = \log_{10} 2 \approx 0.301 \).

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Key Concepts

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

Real-Number Roots
Real-number roots are the solutions of an equation that are real numbers. In the context of quadratic equations, real-number roots indicate the points where the quadratic expression intersects the x-axis. A quadratic equation typically has zero, one, or two real-number roots depending on the value of its discriminant. If the discriminant is positive, as in our problem, the equation has two real-number roots. However, after considering the characteristics of exponential functions in the original exercise, we only take the positive real solution.
Substitution Method
The substitution method is used in solving equations when a transformation simplifies the equation. In the given problem, the substitution of \( y = 10^x \) transformed a more complex exponential equation into a quadratic formula: \( y^2 + 3y - 10 = 0 \). This step is crucial because quadratic equations are often simpler to solve. It involves replacing a more complicated term with a simpler variable, then solving the easier equation. Once solved, the outcome of the simplified equation (\( y \)) is used to find the solution for the original variable \( x \).
Quadratic Formula
The quadratic formula is a reliable tool for finding the roots of any quadratic equation of the form \( ay^2 + by + c = 0 \). The formula is:
  • \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

It helps in determining the solutions without the need for factoring. For our equation \( y^2 + 3y - 10 = 0 \), substitute \( a = 1 \), \( b = 3 \), and \( c = -10 \) into the formula. Simplifying the resulting expression leads us to find the roots. This method is invaluable as it works universally for all quadratic equations, ensuring that at least the fundamental solution is reached.
Discriminant
The discriminant, given by \( b^2 - 4ac \), is a component of the quadratic formula that determines the nature of the roots. In our scenario, the discriminant helps us understand the number and type of solutions we can expect:
  • A positive discriminant indicates two distinct real roots.
  • A zero discriminant signifies exactly one real root.
  • A negative discriminant means there are no real roots, but rather complex ones.

In our equation \( y^2 + 3y - 10 = 0 \), the calculated discriminant is 49, which is positive. Thus, the equation has two real solutions, although due to the exponential nature of the original equation, we only proceed with the viable non-negative solution \( y = 2 \), leading to \( x = \log_{10} 2 \).

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Most popular questions from this chapter

Let \(\mathcal{N}=\mathcal{N}_{0 e^{k t}} .\) In this exercise we show that if \(\Delta t\) is very small, then \(\Delta \mathcal{N} / \Delta t \approx k \mathcal{N} .\) In other words, over very small intervals of time, the average rate of change of \(\mathcal{N}\) is proportional to \(\mathcal{N}\) itself. (a) Show that the average rate of change of the function \(\mathcal{N}=\mathcal{N}_{0} e^{t t}\) on the interval \([t, t+\Delta t]\) is given by $$\frac{\Delta \mathcal{N}}{\Delta t}=\frac{\mathcal{N}_{0} e^{k t}\left(e^{k \Delta t}-1\right)}{\Delta t}=\frac{\mathcal{N}\left(e^{k \Delta t}-1\right)}{\Delta t}$$ (b) In Exercise 26 of Section 5.2 we saw that \(e^{x} \approx x+1\) when \(x\) is close to zero. Thus, if \(\Delta t\) is sufficiently small, we have \(e^{k \Delta t} \approx k \Delta t+1 .\) Use this approximation and the result in part (a) to show that \(\Delta \mathcal{N} / \Delta t \approx k N\) when \(\Delta t\) is sufficiently close to zero.

The intensity of the sounds that the human ear can detect varies over a very wide range of values. For instance, a whisper from 1 meter away has an intensity of approximately \(10^{-10}\) watts per square meter \(\left(\mathrm{W} / \mathrm{m}^{2}\right)\), whereas, from a distance of 50 meters, the intensity of a launch of the Space Shuttle is approximately \(10^{8} \mathrm{W} / \mathrm{m}^{2} .\) For a sound with intensity \(I\), the sound level \(\beta\) is defined by $$ \beta=10 \log _{10}\left(I / I_{0}\right) $$ where the constant \(I_{0}\) is the sound intensity of a barely audible sound at the threshold of hearing. The units for the sound level \(\beta\) are decibels, abbreviated dB. (a) Solve the equation \(\beta=10 \log _{10}\left(1 / I_{0}\right)\) for \(I\) by first dividing by 10 and then converting to exponential form. (b) The sound level for a power lawnmower is \(\beta=100 \mathrm{db}\). and that for a cat purring is \(\beta=10\) db. Use your result in part (a) to determine how many times more intense is the power mower sound than the cat's purring.

Graph each function and specify the domain, range, intercept(s), and asymptote. $$y=\ln (x+e)$$

Solve each equation and solve for \(x\) in terms of the other letters. $$y=A e^{k x}$$

Exercises \(55-60\) introduce a model for population growth that takes into account limitations on food and the environment. This is the logistic growth model, named and studied by the nineteenth century Belgian mathematician and sociologist Pierre Verhulst. (The word "logistic" has Latin and Greek origins meaning "calculation" and "skilled in calculation," respectively. However, that is not why Verhulst named the curve as he did. See Exercise 56 for more about this.) In the logistic model that we "I study, the initial population growth resembles exponential growth. But then, at some point owing perhaps to food or space limitations, the growth slows down and eventually levels off, and the population approaches an equilibrium level. The basic equation that we'll use for logis- tic growth is where \(\mathcal{N}\) is the population at time \(t, P\) is the equilibrium population (or the upper limit for population), and a and b are positive constants. $$\mathcal{N}=\frac{P}{1+a e^{-b t}}$$ The following figure shows the graph of the logistic function \(\mathcal{N}(t)=4 /\left(1+8 e^{-t}\right) .\) Note that in this equation the equilibrium population \(P\) is 4 and that this corresponds to the asymptote \(\mathcal{N}=4\) in the graph. (a) Use the graph and your calculator to complete the following table. For the values that you read from the graph, estimate to the nearest \(0.25 .\) For the calculator values, round to three decimal places. (b) As indicated in the graph, the line \(\mathcal{N}=4\) appears to be an asymptote for the curve. Confirm this empirically by computing \(\mathcal{N}(10), \mathcal{N}(15),\) and \(\mathcal{N}(20) .\) Round each answer to eight decimal places. (c) Use the graph to estimate, to the nearest integer, the value of \(t\) for which \(\mathcal{N}(t)=3\) (d) Find the exact value of \(t\) for which \(\mathcal{N}(t)=3 .\) Evaluate the answer using a calculator, and check that it is consistent with the result in part (c). TABLE AND GRAPH CANT COPY
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