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Chapter 5: Problem 20

Solve each equation for \(x\) by converting to exponential form. In Exercises \(19(b)\) and \(20,\) give two forms for each answer: one involving e and the other a calculator approximation rounded to two decimal places. (a) \(\log _{5} x=e\) (b) \(\ln x=-e\)

Short Answer

Expert verified
(a) \( x = 5^e \approx 79.43 \), (b) \( x = e^{-e} \approx 0.07 \).

Step by step solution

01

Convert to Exponential Form for (a)

The equation given is \( \log_{5} x = e \). The logarithmic form \( \log_{b} a = c \) is equivalent to the exponential form \( a = b^c \). Applying this conversion, \( x = 5^e \).
02

Convert to Exponential Form for (b)

For the equation \( \ln x = -e \), recall that the natural logarithm \( \ln x \) means \( \log_e x \). The exponential form of this is \( x = e^{-e} \).
03

Calculate Exponential Approximation for (a)

To find a calculator approximation for \( 5^e \), use a calculator to compute the value. \( 5^e \approx 79.43 \).
04

Calculate Exponential Approximation for (b)

To find a calculator approximation for \( e^{-e} \), compute using a calculator. \( e^{-e} \approx 0.07 \).

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Key Concepts

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

Logarithms
Logarithms are a mathematical concept used to solve for exponents. If you encounter a logarithmic equation like \( \log_b a = c \), it means that the base \( b \) raised to the power of \( c \) equals \( a \). This is a powerful tool because it allows us to work backwards from the result to find what exponent was used. For instance, \( \log_5 x = e \) can be thought of as asking what power you would raise 5 to, in order to get \( x \). Knowing how to convert logarithmic equations to their exponential form is essential for solving them. This conversion makes it easier to understand and work with the numbers involved in the equation.
Natural Logarithm
The natural logarithm is a special type of logarithm where the base is the number \( e \), an irrational and transcendental number approximately equal to 2.71828. When you see \( \ln x \), you're looking at a logarithm with this unique base. The equation \( \ln x = -e \) translates to \( \log_e x = -e \). In exponential form, this becomes \( x = e^{-e} \). This is an example of how natural logarithms are often used in mathematics, especially in areas involving growth rates, scales, and decay because the constant \( e \) appears naturally in many real-world scenarios.
Calculator Approximation
Calculator approximations are crucial when dealing with irrational numbers or complex equations, as they let you work with practical and understandable figures. While \( 5^e \) and \( e^{-e} \) are exact expressions, their decimal approximations are much simpler: \( 5^e \approx 79.43 \) and \( e^{-e} \approx 0.07 \). Most scientific calculators are capable of finding these values to a decimal point, thus making it easier to work with them in practical or everyday situations. Calculators help bridge the gap between abstract mathematical expressions and tangible numbers, especially when approximations to two decimal places are needed in exercises or real-life applications.
Exponents
Exponents are a way of representing repeated multiplication of a base number. For instance, saying \( 5^e \) means that the number 5 is multiplied by itself \( e \) times. In the exercise, you've converted a logarithmic expression into an equation where the unknown can be solved using exponents. Exponents are not always whole numbers. Sometimes, as with \( e^{-e} \), they can be negative, which indicates reciprocal actions like division. Understanding how exponents function helps in tasks such as converting logarithms to their exponential form, which in turn is a valuable skill for algebra, calculus, and beyond.

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Most popular questions from this chapter

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).

Solve the inequality \(\log _{2} x+\log _{2}(x+1)-\log _{2}(2 x+6)<0.\)

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). $$e^{2 t+3}=10$$

Use the following information on \(p H\) Chemists define pH by the formula pH \(=-\log _{10}\left[\mathrm{H}^{+}\right],\) where [H \(^{+}\) ] is the hydrogen ion concentration measured in moles per liter. For example, if \(\left[\mathrm{H}^{+}\right]=10^{-5},\) then \(p H=5 .\) Solutions with \(a\) pH of 7 are said to be neutral; a p \(H\) below 7 indicates an acid: and a pH above 7 indicates a base. (A calculator is helpful for Exercises 49 and 50.1 A chemist adds some acid to a solution changing the \(\mathrm{pH}\) from 6 to \(4 .\) By what factor does the hydrogen ion concentration change? Note: Lower pH corresponds to higher hydrogen ion concentration.
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