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Chapter 5: Problem 34

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. $$\log _{2}(x+4)=2-\log _{2}(x+1)$$

Short Answer

Expert verified
The exact real root is \(x = 0\) and the approximate value is also 0.

Step by step solution

01

Apply Logarithmic Properties

The given equation is \(\log_2(x+4) = 2 - \log_2(x+1)\). We can use the logarithmic property that states if we have \(a = b - c\), it implies \(a + c = b\). Thus, add \(\log_2(x+1)\) to both sides to get:\[\log_2(x+4) + \log_2(x+1) = 2\]
02

Combine Logarithms

Apply the logarithmic property \(\log_b(M) + \log_b(N) = \log_b(M \cdot N)\) to combine the logs on the left side:\[\log_2((x+4)(x+1)) = 2\]
03

Convert Logarithmic Equation to Exponential Form

Recall that \(\log_b(A) = n\) is equivalent to \(b^n = A\). Use this property to convert the logarithmic equation:\[(x+4)(x+1) = 2^2\]Simplifying, we find:\[(x+4)(x+1) = 4\]
04

Expand and Simplify the Equation

Expand the left side and simplify the quadratic equation:\[(x+4)(x+1) = x^2 + x + 4x + 4 = x^2 + 5x + 4\]Set the equation equal to 4 and simplify:\[x^2 + 5x + 4 = 4\]Subtract 4 from both sides:\[x^2 + 5x = 0\]
05

Solve the Quadratic Equation

Factor the quadratic equation by taking out the common term:\[x(x + 5) = 0\]Set each factor equal to zero and solve for \(x\):1. \(x = 0\)2. \(x + 5 = 0\) which gives \(x = -5\)
06

Verify the Roots

Substitute the calculated values back into the original equation to check their validity:- For \(x = 0\): \(\log_2(0+4) = 2-\log_2(0+1)\) becomes \(2 = 2\), valid.- For \(x = -5\): Substitute into \(\log_2(-5+4)\), which is not defined as the argument \(-1\) is negative.Thus, \(x = -5\) is not a valid root.

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Key Concepts

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

Logarithmic Properties
Logarithmic properties are essential tools in solving equations involving logarithms. Logs help us deal with exponential forms by converting multiplicative relationships into additive ones. When you encounter equations like \( \log_b(M \cdot N) = \log_b(M) + \log_b(N) \), it simplifies multiplication inside a log to a simple addition outside.
We also use the property \( \log_b(M) = n \) translates to \(b^n = M\). This conversion is crucial when simplifying equations like the one in the exercise. These fundamental properties make it easier to manipulate and solve complex equations that would otherwise be cumbersome to handle.
  • Addition property: Combines logs with the same base.
  • Subtraction property: Allows expressing differences as divisions.
  • Change of base formula: Helps solve equations with uncommon bases.
Utilizing these properties correctly aids in solving logarithmic equations to find their real-number roots.
Quadratic Equation
A quadratic equation is a polynomial equation of degree two. Its general form is \(ax^2 + bx + c = 0\). In the exercise, we have \(x^2 + 5x = 0\).
You can solve quadratic equations through various methods:
  • Factoring.
  • Completing the square.
  • Quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

This method of factoring shows that the quadratic can be written as a product of linear factors. Solving \(x(x + 5) = 0\) results in roots \(x = 0\) or \(x = -5\). Just like in the problem, it shows how efficiently quadratic equations can be handled when they are set up and factored correctly.
Exponential Form
Exponential form is a way of expressing equations that involve powers or exponents. When solving logarithmic equations, converting them to exponential form can sometimes make them easier to solve. For instance, the equation \(\log_2((x+4)(x+1)) = 2\) can be converted to its exponential form \((x+4)(x+1) = 2^2\).

This conversion translates the log equation into a multiplication problem, allowing us to solve for real-number roots using algebraic techniques like expansion. It effectively bridges the gap between logarithmic operations and polynomial equations. Understanding exponential form is key to simplifying and solving problems involving logarithms and leads smoothly into solving for roots.
Validity of Solutions
Checking the validity of solutions is a critical step in solving equations, especially involving logarithms. Real-number roots must satisfy the condition that the argument of a logarithmic function is positive. For example, when verifying solutions from our quadratic equation:
  • For \(x = 0 \), the equation is satisfied because \(\log_2(4) = 2\).
  • For \(x = -5\), a problem arises as \(\log_2(-1)\) is undefined (negative arguments are not in a logarithm's domain).

Ensuring the arguments of log functions result in valid real numbers is crucial. Solutions must be within the domain of the functions involved or they aren't considered valid. Double-checking ensures you do not carry forward incorrect solutions due to simple domain issues.

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Most popular questions from this chapter

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

(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). $$\left(10^{x}\right)^{2}=40$$

The Chernobyl nuclear explosion (in the former Soviet Union, on April 26,1986 ) released large amounts of radioactive substances into the atmosphere. These substances included cesium-137, iodine-131, and strontium-90. Although the radioactive material covered many countries, the actual amount and intensity of the fallout varied greatly from country to country, due to vagaries of the weather and the winds. One area that was particularly hard hit was Lapland, where heavy rainfall occurred just when the Chernobyl cloud was overhead. (a) Many of the pastures in Lapland were contaminated with cesium-137, a radioactive substance with a half- life of 33 years. If the amount of cesium- 137 was found to be ten times the normal level, how long would it take until the level returned to normal? Hint: Let \(\mathcal{N}_{0}\) be the amount that is ten times the normal level. Then you want to find the time when \(\mathcal{N}(t)=\mathcal{N}_{0} / 10\) (b) Follow part (a), but assume that the amount of cesium-137 was 100 times the normal level. Remark: Several days after the explosion, it was reported that the level of cesium- 137 in the air over Sweden was 10,000 times the normal level. Fortunately there was little or no rainfall.

Simplify \(a^{x}\) when \(x=\log _{b}\left(\log _{b} a\right) / \log _{b} a\)

(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). $$10^{2 x-1}=145$$
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