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Chapter 5: Problem 41

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. $$\log _{10}(x-6)+\log _{10}(x+3)=1$$

Short Answer

Expert verified
The real root is \( x = 7 \).

Step by step solution

01

Apply Logarithm Property

First, we use the property of logarithms: \( \log_b(A) + \log_b(B) = \log_b(AB) \). Rewrite the equation: \( \log_{10}((x-6)(x+3)) = 1 \).
02

Convert Logarithm to Exponential Form

To eliminate the logarithm, convert the equation to its exponential form using the definition \( \log_b(A) = C \implies b^C = A \). Here, \( 10^1 = (x-6)(x+3) \).
03

Expand and Simplify the Equation

Simplify \( 10 = (x-6)(x+3) \) by expanding the right side: \( x^2 + 3x - 6x - 18 = x^2 - 3x - 18 \). The equation becomes \( x^2 - 3x - 18 = 10 \).
04

Rearrange into a Quadratic Equation

Move all terms to one side to form a standard quadratic equation: \( x^2 - 3x - 28 = 0 \).
05

Solve the Quadratic Equation

Solve the quadratic equation \( x^2 - 3x - 28 = 0 \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = -3 \), and \( c = -28 \).
06

Evaluate the Quadratic Formula

Calculate the discriminant: \( (-3)^2 - 4(1)(-28) = 9 + 112 = 121 \). Now apply the quadratic formula: \( x = \frac{3 \pm \sqrt{121}}{2} \).
07

Find Exact Solutions

The square root of 121 is 11, so \( x = \frac{3 + 11}{2} = 7 \) and \( x = \frac{3 - 11}{2} = -4 \).
08

Discard Extraneous Solutions

Verify which solutions are valid in the original equation's domain. Since \( \log_{10}(x-6) \) and \( \log_{10}(x+3) \) require positive arguments, valid solutions must satisfy \( x > 6 \). Hence, discard \( x = -4 \).
09

Provide Calculator Approximation

For this exercise, no approximation is necessary since the exact root, \( x = 7 \), is an integer.

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Key Concepts

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

Logarithm Properties
When solving equations involving logarithms, it's essential to leverage the properties of logarithms to simplify expressions. One of the most useful properties is the product property:
  • \( \log_b(A) + \log_b(B) = \log_b(AB) \).
This property allows you to combine two logarithms with the same base into a single logarithm. In our exercise:
  • \( \log_{10}(x-6) + \log_{10}(x+3) = \log_{10}((x-6)(x+3)) \).
By using this property, we have rewritten the original equation in a more manageable form. This simplification is crucial as it sets up the equation for further transformations and solving. It reduces the complexity by condensing the problem to a single logarithmic expression, which can then be converted into exponential form.
Quadratic Equation
A quadratic equation is any equation that can be written in the standard form: \( ax^2 + bx + c = 0 \). In our problem, after simplifying and using logarithm properties:
  • The quadratic equation we get is \( x^2 - 3x - 28 = 0 \).
Quadratic equations are polynomials of degree two, and they are characterized by having up to two real solutions.

To solve these types of equations, we have different methods:
  • Factoring (if possible)
  • Completing the square
  • Using the quadratic formula
In many cases, such as this exercise, the quadratic formula is the go-to method, especially when the quadratic does not factor easily. Moreover, quadratic equations are an essential part of algebra because they appear in various real-world contexts.
Exponential Form
Converting from logarithmic to exponential form is a major step in solving equations that involve logarithms. The basic idea is:
  • If \( \log_b(A) = C \), then \( b^C = A \).
This conversion eliminates the logarithm, allowing one to solve for the variable inside. In our example:
  • \( \log_{10}((x-6)(x+3)) = 1 \) transforms to \( 10^1 = (x-6)(x+3) \).
This exponential form \( 10 = (x-6)(x+3) \) reflects the power of understanding logarithms. It translates an equation of logarithmic nature to a polynomial form that's more straightforward to handle. This transformation is crucial for progressing to solving real-number roots.
Quadratic Formula
The quadratic formula is a tool that allows you to find the roots of any quadratic equation. The formula is:
  • \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
The discriminant, \( b^2 - 4ac \), determines the nature of the roots:
  • If positive, there are two distinct real roots.
  • If zero, there's exactly one real root (a repeated root).
  • If negative, the roots are complex and not real.
For our quadratic equation \( x^2 - 3x - 28 = 0 \), the discriminant is 121, meaning two distinct real roots exist.

Applying the quadratic formula:
  • \( x = \frac{3 \pm 11}{2} \).
Yields the roots \( x = 7 \) and \( x = -4 \). However, due to the logarithmic domain constraints, only \( x = 7 \) is valid. The quadratic formula thereby not only provides the potential solutions but also underscores the need to consider the domain of the original equation carefully.

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Most popular questions from this chapter

The following extract is from an article by Kim Murphy that appeared in the Los Angeles Times on September 14 1994 CAIRO-Over a chorus of reservations from Latin America and Islamic countries still troubled about abortion and family issues, nearly 180 nations adopted a wide-ranging plan Tuesday on global population, the first in history to obtain partial endorsement from the Vatican. The plan, approved on the final day of the U.N. population conference here, for the first time tries to limit the growth of the world's population by preventing it from exceeding 7.2 billion people over the next two decades. (a) In 1995 the world population was 5.7 billion, with a relative growth rate of \(1.6 \% /\) year. Assuming continued exponential growth at this rate, make a projection for the world population in the year \(2020 .\) Round off the answer to one decimal place. How does your answer compare to the target value of 7.2 billion mentioned in the article? (b) As in part (a), assume that in 1995 the world population was 5.7 billion. Determine a value for the growth constant \(k\) so that exponential growth throughout the years \(1995-2020\) leads to a world population of 7.2 billion in the year 2020 .

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$e^{2+x} \geq 100$$

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

(a) Use a graphing utility to estimate the root(s) of the equation to the nearest one-tenth (as in Example 6). (b) Solve the given equation algebraically by first rewriting it in logarithmic form. Give two forms for each answer: an exact expression and a calculator approximation rounded to three decimal places. Check to see that each result is consistent with the graphical estimate obtained in part (a). $$10^{2 x-1}=145$$

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. $$\mathcal{N}=\frac{P}{1+a e^{-b t}}$$ (Continuation of Exercise 55 ) The author's ideas for this exercise are based on Professor Bonnie Shulman's article "Math-Alive! Using Original Sources to Teach Mathematics in Social Context," Primus, vol. VIII (March \(1998)\) (a) The function \(\mathcal{N}\) in Exercise 55 expresses population as a function of time. But as pointed out by Professor Shulman, in Verhulst's original work it was the other way around; he expressed time as a function of population. In terms of our notation, we would say that he was studying the function \(\mathcal{N}^{-1}\). Given \(\mathcal{N}(t)=4 /\left(1+8 e^{-t}\right)\) find \(\mathcal{N}^{-1}(t)\) (b) Use a graphing utility to draw the graphs of \(\mathcal{N}, \mathcal{N}^{-1}\), and the line \(y=x\) in the viewing rectangle [-3,8,2] by \([-3,8,2] .\) Use true portions. (Why?) (c) In the viewing rectangle [0,5,1] by \([-3,2,1],\) draw the graphs of \(y=\mathcal{N}^{-1}(t)\) and \(y=\ln t .\) Note that the two graphs have the same general shape and characteristics. In other words, Verhulst's logistic function (our \(\mathcal{N}^{-1}\) ) appears log-like, or logistique, as Verhulst actually named it in French. (For details, both historical and mathematical, see the paper by Professor Shulman cited above.)
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