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(a) How large must \(x\) be before the graph of \(y=\ln x\) reaches a height of \(y=100 ?\) (b) How large must \(x\) be before the graph of \(y=e^{x}\) reaches a height of (i) \(y=100 ?\) (ii) \(y=10^{6} ?\)

Short Answer

Expert verified
(a) \(x = e^{100}\); (b)(i) \(x = \ln(100)\); (ii) \(x = 6\cdot\ln(10)\).

Step by step solution

01

Understanding the Problem

We need to find the value of \(x\) for two different logarithmic and exponential equations. First, we solve for \(x\) in \(y = \ln x\) when \(y = 100\). Next, we solve for \(x\) in \(y = e^x\) for two cases: when \(y = 100\) and \(y = 10^6\).
02

Solving part (a)

For the function \(y = \ln x\), we find the value of \(x\) when \(y = 100\). This means solving the equation \(\ln x = 100\).To find \(x\), exponentiate both sides: \[ x = e^{100} \].Thus, \(x\) must be \(e^{100}\) for the graph to reach a height of 100.
03

Solving part (b)(i)

For the function \(y = e^x\), we want to find \(x\) when \(y = 100\). Set up the equation:\[ e^x = 100 \].Taking the natural logarithm of both sides gives:\[ x = \ln(100) \].Calculate \(\ln(100)\) to find \(x\).
04

Solving part (b)(ii)

For the function \(y = e^x\), we need to find \(x\) when \(y = 10^6\). Set the equation:\[ e^x = 10^6 \].Taking the natural logarithm of both sides results in:\[ x = \ln(10^6) \].Use the logarithm rule \(\ln(10^6) = 6 \cdot \ln(10)\) to compute \(x\).

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Key Concepts

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

Natural Logarithms
Natural logarithms are an essential concept in understanding exponential and logarithmic functions. They help us solve problems involving growth and decay. A natural logarithm is the logarithm to the base of Euler's number, denoted as "e", which is approximately 2.718281828. In mathematical notation, the natural logarithm of a number x is written as \(\ln(x)\).

Key characteristics of natural logarithms include:
  • They grow very slowly compared to exponential functions.
  • The natural logarithm of 1 is 0, i.e., \(\ln(1) = 0\).
  • As x increases, \(\ln(x)\) increases but at a decreasing rate.
In our problem, we were tasked to find the value of \(x\) for which the graph of \(y = \ln x\) would reach a height of 100. By setting up the equation \(\ln x = 100\), we used the fact that exponentiation is the inverse operation to logarithms. Hence, the solution involved calculating \(x = e^{100}\). This points out the extensive growth rate of the function despite the slow increase in the log values relative to the input size.
Exponential Growth
Exponential growth is a process where the quantity increases at a rate proportional to its current value. This characteristic leads to faster and faster growth as time progresses.

Some critical points about exponential growth are:
  • When graphed, exponential growth results in a J-shaped curve that shows accelerated growth.
  • In contrast to linear functions, the rate of change in exponential functions continuously increases.
  • Common in scenarios like population growth, radioactive decay, and financial investments.
In the exercise, we explored exponential functions through \(y = e^x\). For part (b)(i), we found \(x\) with \(y = 100\) by setting \(e^x = 100\) and solving for \(x = \ln(100)\). Similarly, part (b)(ii) required finding \(x\) for \(y = 10^6\), leading us to \(x = \ln(10^6)\). Using the logarithm property \(\ln(10^6) = 6 \cdot \ln(10)\), we determine how quickly exponential growth overtakes linear or even polynomial growth.
Solving Equations
Solving equations involving exponential and logarithmic functions demands understanding each function's properties and relationships. Here, solving involved the use of natural logarithms and exponential functions interchangeably due to their inverse nature.

Steps to solve these equations effectively include:
  • Identify the type of function involved (exponential or logarithmic).
  • Apply the inverse operation: use the natural log (\(\ln\)) to solve exponential equations and exponentiate to solve logarithmic equations.
  • Simplify the equations using basic logarithmic properties, like \(\ln(a^b) = b \cdot \ln(a)\).
For the \(\ln x = 100\) equation in part (a), solving was straightforward by exponentiating both sides. In part (b), applying the natural log allowed us to convert equations \(e^x = 100\) and \(e^x = 10^6\) into simpler logarithmic equations, highlighting how these mathematical strategies aid in demystifying complex formulae into manageable steps.

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Most popular questions from this chapter

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.

(a) Suppose that a certain country violates the ban against above-ground nuclear testing and, as a result, an island is contaminated with debris containing the radioactive substance iodine-131. A team of scientists from the United Nations wants to visit the island to look for clues in determining which country was involved. However, the level of radioactivity from the iodine- 131 is estimated to be 30,000 times the safe level. Approximately how long must the team wait before it is safe to visit the island? The half-life of iodine- 131 is 8 days. (b) Rework part (a), assuming instead that the radioactive substance is strontium-90 rather than iodine-131. The half-life of strontium- 90 is 28 years. Assume, as before, that the initial level of radioactivity is 30,000 times the safe level. (This exercise underscores the difference between a half-life of 8 days and one of 28 years.)

Decide which of the following properties apply to each function. (More than one property may apply to a function.)A. The function is increasing for \(-\infty

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

(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). $$e^{3 x^{2}}=112$$

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