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Chapter 5: Problem 91

$$\text { The given equations are quadratic in form. Solve each and give exact solutions.}$$ $$(\ln x)^{2}+16=10 \ln x$$

Short Answer

Expert verified
The exact solutions are \( x = e^8 \) and \( x = e^2 \).

Step by step solution

01

Recognize the Quadratic Form

Identify that the equation \((\ln x)^{2} + 16 = 10 \ln x\) is quadratic in form, but with \( \ln x \) as the variable. Rewriting it gives us \((\ln x)^{2} - 10 \ln x + 16 = 0\).
02

Use Substitution

Substitute \( y = \ln x \) into the equation. The equation becomes \( y^2 - 10y + 16 = 0 \).
03

Solve the Quadratic Equation

Apply the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1, b = -10, c = 16 \). Calculate the discriminant: \( b^2 - 4ac = (-10)^2 - 4 \cdot 1 \cdot 16 = 100 - 64 = 36 \).
04

Apply the Quadratic Formula

Plug the values into the quadratic formula: \( y = \frac{10 \pm \sqrt{36}}{2} \). This simplifies to \( y = \frac{10 \pm 6}{2} \). Thus, the solutions are \( y = \frac{16}{2} = 8 \) and \( y = \frac{4}{2} = 2 \).
05

Reverse the Substitution

Since \( y = \ln x \), substitute back to find \( x \). For \( y = 8 \), \( \ln x = 8 \) so \( x = e^8 \). For \( y = 2 \), \( \ln x = 2 \) so \( x = e^2 \).
06

Write the Solution

The exact solutions in terms of \( x \) are \( x = e^8 \) and \( x = e^2 \).

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Key Concepts

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

Logarithms
Logarithms serve as a crucial tool in mathematics to deal with exponential relationships. They help transform multiplicative processes into additive ones, simplifying complex calculations. The logarithm of a number is the exponent to which another fixed number, the base, must be raised to produce that number. In the equation \((\ln x)^2 + 16 = 10 \ln x\), \(\ln x\) represents the natural logarithm of \(x\), which uses the base \(e\). The natural logarithm simplifies the operations on exponential equations, making them easier to handle when dealing with quadratic forms.
  • The natural logarithm (\(\ln(x)\)) is the inverse operation to exponentiation with base \(e\).
  • If \(y = \ln(x)\), then this means \(x = e^y\).
  • In the context of solving equations, logarithms allow us to revert back to the original variable once the quadratic form has been solved.
Understanding logarithms is key to dealing with equations where the variable appears in an exponential form.
Substitution Method
The substitution method is a strategy used to simplify solving equations by reducing them into a more familiar form. In this exercise, we replaced the variable \(\ln x\) with \(y\) to transform the equation from \((\ln x)^2 + 16 = 10 \ln x\) into a simpler quadratic, \(y^2 - 10y + 16 = 0\).
  • Substitution is especially useful when dealing with non-linear terms as it can turn them into linear ones.
  • By substituting, you effectively create a new equation in terms of a single variable, simplifying the resolution process.
  • After solving in terms of the new variable, you must remember to substitute back to find the solutions in terms of the original variable.
This method nicely bridges intermediate and advanced algebra, making the problem significantly more straightforward to tackle.
Quadratic Formula
The quadratic formula is a powerful tool for finding the roots of any quadratic equation. It is particularly useful when factorization is difficult or impossible. For a quadratic equation of the form \(ax^2 + bx + c = 0\), the quadratic formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In our problem, we applied this formula to \(y^2 - 10y + 16 = 0\) with \(a = 1\), \(b = -10\), and \(c = 16\). After calculating the discriminant \(b^2 - 4ac\), which was \(36\), we computed the solutions:
  • \( y = \frac{10 + 6}{2} = 8\)
  • \( y = \frac{10 - 6}{2} = 2\)
The quadratic formula gives exact solutions to the equation, making it an indispensable tool in your algebra toolkit.
Exact Solutions
Exact solutions provide a precise answer to mathematical problems without approximations. When using exact methods, like employing logarithms and quadratic equations, we retain the full accuracy of the mathematical process. In solving \((\ln x)^2 + 16 = 10 \ln x\), exact solutions were achieved at \(x = e^8\) and \(x = e^2\), owing to the exactness of both the exponential function and logarithms.
  • Exact solutions allow mathematicians and scientists to maintain precision in calculations.
  • This precision becomes notably important in fields where small deviations can lead to large errors, like engineering and computer science.
  • Unlike numerical approximations, exact solutions give the most reliable long-term results.
Understanding and performing computations for exact solutions is crucial to mastering higher-level mathematics, offering complete clarity and accuracy.

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Most popular questions from this chapter

Emissions Governments could reduce carbon emissions by placing a tax on fossil fuels. The costbenefit equation $$\ln (1-P)=-0.0034-0.0053 x$$ estimates the relationship between a tax of \(x\) dollars per ton of carbon and the percent \(P\) reduction in emissions of carbon, where \(P\) is in decimal form. Determine \(P\) when \(x=60 .\) Interpret the result. (Source: Clime, W., The Economics of Global Warming, Institute for International Economics.)

In the formula \(A=P\left(1+\frac{r}{n}\right)^{n t},\) we can interpret \(P\) as the present value of A dollars t years from now, earning annual interest \(r\) compounded \(n\) times per year. In this context, \(A\) is called the future value. If we solve the formula for \(P,\) we obtain $$P=A\left(1+\frac{r}{n}\right)^{-n t}$$ Use the future value formula. Find the present value of an account that will be worth \(\$ 25,000\) in 2.75 years, if interest is compounded quarterly at \(6 \%\).

In the formula \(A=P\left(1+\frac{r}{n}\right)^{n t},\) we can interpret \(P\) as the present value of A dollars t years from now, earning annual interest \(r\) compounded \(n\) times per year. In this context, \(A\) is called the future value. If we solve the formula for \(P,\) we obtain $$P=A\left(1+\frac{r}{n}\right)^{-n t}$$ Use the future value formula. Estimate the interest rate necessary for a present value of \(\$ 1200\) to grow to a future value of \(\$ 1408\) if interest is compounded quarterly for 8 years.

The concentration of bacteria \(B\) in millions per milliliter after \(x\) hours is given by $$B(x)=1.33 e^{0.15 x}$$ (a) How many bacteria are there after 2.5 hours? (b) How many bacteria are there after 8 hours? (c) After how many hours will there be 31 million bacteria per milliliter?

In 1666 the village of Eyam, located in England, experienced an outbreak of the Great Plague. Out of 261 people in the community, only 83 survived. The table shows a function \(f\) that computes the number of people who had not (yet) been infected after \(x\) days. $$\begin{array}{|c|r|r|r|r|}\hline x & 0 & 15 & 30 & 45 \\\\\hline f(x) & 254 & 240 & 204 & 150 \\\\\hline\end{array}$$ $$\begin{array}{|c|c|c|c|c|}\hline x & 60 & 75 & 90 & 125 \\\\\hline f(x) & 125 & 103 & 97 & 83\\\\\hline\end{array}$$ (a) Use a table to represent a function \(g\) that computes the number of people in Eyam who were infected after \(x\) days. (b) Write an equation that shows the relationship between \(f(x)\) and \(g(x)\) (c) Use graphing to decide which equation represents \(g(x)\) better \(y_{1}=\frac{171}{1+18.6 e^{-0.0747 x}}\) or \(y_{2}=18.3(1.024)^{x}\) (d) Use your results from parts (b) and (c) to find a formula for \(f(x)\)
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