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Chapter 5: Problem 35

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

Short Answer

Expert verified
The roots of the equation are approximately \( x \approx \pm 1.265 \).

Step by step solution

01

Use a Graphing Utility

Using a graphing utility or a calculator, plot the function \( y = 10^{x^2} \) and find where it intersects with the horizontal line \( y = 40 \). The intersection points will approximate the values of \( x \) where \( 10^{x^2} = 40 \). Estimate these points to the nearest one-tenth, which will help us determine the possible roots for the equation.
02

Rewrite in Logarithmic Form

To solve the equation algebraically, first take the logarithm of both sides of the equation \( 10^{x^2} = 40 \). You can use the common logarithm (base 10) for simplicity: \[\log_{10}(10^{x^2}) = \log_{10}(40)\] Using the property of logarithms \( \log_{10}(a^b) = b \cdot \log_{10}(a) \), the equation becomes:\[ x^2 = \log_{10}(40)\]
03

Solve for x

Now solve for \( x \) by taking the square root of both sides:\[x = \pm \sqrt{\log_{10}(40)}\]This expression gives the exact solutions for \( x \).
04

Calculator Approximation

Use a calculator to find a decimal approximation for \( \log_{10}(40) \) and then compute \( x = \pm \sqrt{\log_{10}(40)}\). Calculate:\[ \log_{10}(40) \approx 1.602 \]Therefore, the approximate solutions are:\[ x \approx \pm \sqrt{1.602} \approx \pm 1.265\]
05

Check Consistency

Verify that the approximate solutions \( x \approx \pm 1.265 \) are consistent with the graphical estimates obtained in Step 1. This can be done by ensuring the graph intersected near these x-values, confirming the calculations matched the graphical result.

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Key Concepts

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

Graphing Utility
A graphing utility is a helpful tool for visualizing mathematical functions. In mathematics, a graphing utility could be a software program or a feature within a calculator that allows you to graph equations and analyze those graphs. Using a graphing utility is one of the first steps in solving complex equations, like the one given in the exercise.
  • To use a graphing utility for the equation \( y = 10^{x^2} \), you would input this function and the constant line \( y = 40 \) into the device.
  • The tool will graph both the curve and the horizontal line, helping us determine where they intersect.
  • The points of intersection between the curve and line show the approximate roots of the equation, which are the values of \( x \) that satisfy \( 10^{x^2} = 40 \).
  • Estimate the intersection points as close as possible, maybe to the nearest one-tenth, to get the values needed for further calculation.
A graphing utility provides a visual check and starting point for calculating more precise numerical values.
Logarithmic Form
Rewriting an equation in logarithmic form is an algebraic technique used to make solving exponential equations easier. Here's how it works for the equation \(10^{x^2} = 40\).
  • To convert the exponential form to logarithmic form, apply the logarithm (base 10) to both sides of the equation.
  • This transforms the equation using the property \( \log_{10}(a^b) = b \log_{10}(a) \, \) which simplifies to \( x^2 = \log_{10}(40)\).
By rewriting in logarithmic form, we simplify the original equation and can solve for \( x \) algebraically. This process makes it easier to find the exact solutions without approximating prematurely.
Calculator Approximation
After finding the expression \(x = \pm \sqrt{\log_{10}(40)}\), using a calculator is necessary to find a numerical approximation. This step is crucial since logarithms often yield decimal results.
  • First, calculate \(\log_{10}(40)\), which roughly equals 1.602. You can do this using a standard scientific calculator or an online calculator.
  • Next, find the square root of this value to find \( x \). This calculation gives \( x \approx \pm \sqrt{1.602} \approx \pm 1.265 \).
  • Ensure your calculator is set to compute to at least three decimal places for accuracy.
This method provides the approximate solutions \( x \approx \pm 1.265 \). These numerical results are consistent with the graphing utility estimations and verify our solutions.
Exact Solutions
While we've approached solving the equation numerically, finding exact solutions is often necessary for a complete understanding. Exact solutions go beyond approximations and offer precise results that can be used in further calculations or proofs.
  • The exact form of the solution after rewriting in logarithmic form was \(x = \pm \sqrt{\log_{10}(40)}\).
  • This expresses the solution in terms of the logarithmic function, providing insight into the structure of the equation without rounding off any values prematurely.
  • Exact solutions are critical as they can help confirm the results obtained through approximations and ensure there's consistency in problem-solving.
Maintaining exactness in solutions is key in higher mathematics, where precision is often demanded for deeper analysis and accuracy.

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Most popular questions from this chapter

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$\ln (2-5 x)>2$$

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

Strontium-90, with a half-life of 28 years, is a radioactive waste product from nuclear fission reactors. One of the reasons great care is taken in the storage and disposal of this substance stems from the fact that strontium-90 is, in some chemical respects, similar to ordinary calcium. Thus strontium-90 in the biosphere, entering the food chain via plants or animals, would eventually be absorbed into our bones. (a) Compute the decay constant \(k\) for strontium-90. (b) Compute the time required if a given quantity of strontium-90 is to be stored until the radioactivity is reduced by a factor of \(1000 .\) (c) Using half-lives, estimate the time required for a given sample to be reduced by a factor of \(1000 .\) Compare your answer with that obtained in (b).

The half-life of plutonium-241 is 13 years. (a) How much of an initial 2 -g sample remains after 5 years? (b) Find the time required for \(90 \%\) of the 2 -g sample to decay. Hint: If \(90 \%\) has decayed, then \(10 \%\) remains.

The age of some rocks can be estimated by measuring the ratio of the amounts of certain chemical elements within the rock. The method known as the rubidium-strontium method will be discussed here. This method has been used in dating the moon rocks brought back on the Apollo missions. Rubidium-87 is a radioactive substance with a half-life of \(4.7 \times 10^{10}\) years. Rubidium- 87 decays into the substance strontium- \(87,\) which is stable (nonradioactive). We are going to derive the following formula for the age of a rock: $$T=\frac{\ln \left[\left(\mathcal{N}_{s} / \mathcal{N}_{r}\right)+1\right]}{-k}$$ where \(T\) is the age of the rock, \(k\) is the decay constant for rubidium-87, \(\mathcal{N}_{s}\) is the number of atoms of strontium-87 now present in the rock, and \(\mathcal{N},\) is the number of atoms of rubidium-87 now present in the rock. (a) Assume that initially, when the rock was formed, there were \(\mathcal{N}_{0}\) atoms of rubidium-87 and none of strontium-87. Then, as time goes by, some of the rubidium atoms decay into strontium atoms, but the to tal number of atoms must still be \(\mathcal{N}_{0} .\) Thus, after \(T\) years, we have \(\mathcal{N}_{0}=\mathcal{N}_{r}+\mathcal{N}_{s}\) or, equivalently, $$ \mathcal{N}_{s}=\mathcal{N}_{0}-\mathcal{N}_{r}$$However, according to the law of exponential decay for the rubidium-87, we must have \(\mathcal{N}_{r}=\mathcal{N}_{0} e^{k T} .\) Solve this equation for \(\mathcal{N}_{0}\) and then use the result to eliminate \(\mathcal{N}_{0}\) from equation \((1) .\) Show that the result can be written $$\mathcal{N}_{s}=\mathcal{N}_{r} e^{-k T}-\mathcal{N}_{r}$$ (b) Solve equation (2) for \(T\) to obtain the formula given at the beginning of this exercise.
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