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Chapter 3: Problem 96

In Exercises \(93-102,\) solve each equation. $$3|\log x|-6=0$$

Short Answer

Expert verified
The solutions to the equation are \(x = 100\) and \(x = 0.01\).

Step by step solution

01

Remove the constant from the equation

This equation can be simplified by adding 6 to both sides, to isolate the absolute value function: \(3|\log x| = 6\).
02

Remove the coefficient of the absolute value

Divide both sides by 3 to isolate the absolute value: \(|\log x| = 2\). This means that \(\log x\) can be equal to 2 or -2, because the absolute value of both positive and negative 2 is 2.
03

Solve the positive logarithmic equation

First, solve \(\log x = 2\). The base of the logarithm is 10 (since it is not written), so this equation is equivalent to: \(x = 10^2\). Solving this gives \(x = 100\). Double-check this solution by plugging it back into the original equation to confirm.
04

Solve the negative logarithmic equation

Next, solve \(\log x = -2\). This is equivalent to \(x = 10^{-2}\). Solving this gives \(x = 0.01\). Double-check this solution by plugging it back into the original equation to confirm.

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Key Concepts

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

Understanding Absolute Value
Absolute value refers to the distance a number is from zero on a number line, regardless of direction. It's essentially a way to express the magnitude of a number without regard to its sign.
  • The absolute value of a positive number is the number itself.
  • The absolute value of a negative number is the positive counterpart of that number.
  • The absolute value of zero is zero.
In equations, absolute value symbols are used to indicate that a number can be either positive or negative while maintaining a specific absolute value. For example, \( |x| = 3 \) implies that \( x \) could be \( 3 \) or \( -3 \.\)
In the context of logarithmic equations, using absolute values means you consider both the positive and negative cases which can result from an absolute operation, ensuring you account for all possible solutions to the equation.
Decoding Logarithms
A logarithm is the opposite of an exponent, similar to how subtraction is the opposite of addition. When you see \( \log_b a = c \), this states that \( b^c = a \). Here, \( b \) is the base, \( a \) is the number you're taking the logarithm of, and \( c \) is the power to which the base is raised to get \( a \).
In typical problems, sometimes the base isn't explicitly written, which often means it's a common logarithm with base 10.
  • For example, \( \log 100 = 2 \) because \( 10^2 = 100 \).
  • If you have \( \log x = 2 \), it translates to \( x = 10^2 \) or \( x = 100 \).
  • In the same vein, \( \log x = -2 \) becomes \( x = 10^{-2} = 0.01 \).
Logarithms help to solve equations where the unknown is an exponent, making them extremely useful in a variety of mathematical and real-world applications. Understanding how to handle logarithms is crucial to solving more complex equations.
Steps to Solving Equations
Solving equations involves finding the values of the unknown variables that make the equation true. Here, we look at equations through three primary steps:
  • Rewrite the equation: Begin by simplifying or rearranging terms to isolate the part of the equation you want to solve. For instance, adding or subtracting terms as shown when moving -6 to isolate the absolute value term.
  • Isolate the term: This often involves dividing by coefficients to further simplify the equation and focusing on obtaining a clear view of the variable(s) at play. For example, dividing by 3 to fully isolate the absolute value function.
  • Solve for each possibility: In absolute value equations or any equations presenting multiple cases, solve for each case independently. Like splitting \( |\log x| = 2 \) into two separate log equations, \( \log x = 2 \) and \( \log x = -2 \).
Each equation might require different strategies, so it's important to stay flexible and methodical. Checking your solutions in the original equation ensures that they are correct and make sense in the given context.

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Most popular questions from this chapter

The formula \(A=25.1 e^{0.0187 t}\) models the population of Texas, \(A\), in millions, \(t\) years after 2010 . a. What was the population of Texas in \(2010 ?\) b. When will the population of Texas reach 28 million?

From 1970 through \(2010 .\) The data are shown again in the table. Use all five data points to solve Exercises \(70-74\). $$\begin{array}{cc}\hline \begin{array}{c}x, \text { Number of Years } \\\\\text { after } 1969 \end{array} & \begin{array}{c}y, \text { U.S. Population } \\\\\text { (millions) }\end{array} \\ \hline 1(1970) & 203.3 \\\11(1980) & 226.5 \\\21(1990) & 248.7 \\\31(2000) & 281.4 \\\41(2010) & 308.7 \end{array}$$ Use your graphing utility's logarithmic regression option to obtain a model of the form \(y=a+b \ln x\) that fits the data. How well does the correlation coefficient, \(r,\) indicate that the model fits the data?

Solve each equation in Exercises \(146-148 .\) Check each proposed solution by direct substitution or with a graphing utility. $$(\log x)(2 \log x+1)=6$$

The formula \(S=C(1+r)^{t}\) models inflation, where \(C=\) the value today, \(r=\)the annual inflation rate, and \(S=\)the inflated value t years from now. Use this formula to solve. Round answers to the nearest dollar. If the inflation rate is \(3 \%,\) how much will a house now worth \(\$ 510,000\) be worth in 5 years?

Use a calculator with an \(\left[e^{x}\right]\) key to solve. The bar graph shows the percentage of U.S. high school seniors who applied to more than three colleges for selected years from 1980 through 2013. (BAR GRAPH CAN'T COPY) The data can be modeled by $$ f(x)=x+31 \text { and } g(x)=32.7 e^{0.0217 x} $$ in which \(f(x)\) and \(g(x)\) represent the percentage of high school seniors who applied to more than three colleges \(x\) years after 1980\. Use these functions to solve . Where necessary, round answers to the nearest percent. In college, we study large volumes of information \(-\) information that, unfortunately, we do not often retain for very long. The function $$ f(x)=80 e^{-0.5 x}+20 $$ describes the percentage of information, \(f(x),\) that a particular person remembers \(x\) weeks after learning the information. a. Substitute 0 for \(x\) and, without using a calculator, find the percentage of information remembered at the moment it is first learned. b. Substitute 1 for \(x\) and find the percentage of information that is remembered after 1 week. c. Find the percentage of information that is remembered after 4 weeks. d. Find the percentage of information that is remembered after one year (52 weeks).
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