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Chapter 3: Problem 91

Solve \(P=P_{0} e^{k t}\) for \(t\)

Short Answer

Expert verified
The solution is \(t = \frac{\text{ln}\bigg(\frac{P}{P_0}\bigg)}{k}\).

Step by step solution

01

- Isolate the exponential term

Start by isolating the exponential term on one side of the equation. To do this, divide both sides of the equation by the initial value, \(P_0\). This gives: \[\frac{P}{P_0} = e^{kt}\]
02

- Apply the natural logarithm

To solve for \(t\), take the natural logarithm (ln) of both sides of the equation. The purpose of taking the natural logarithm is to get rid of the exponential. This gives: \[\text{ln}\bigg(\frac{P}{P_0}\bigg) = \text{ln}(e^{kt})\]
03

- Simplify the equation

Use the property of logarithms that \(\text{ln}(e^{x}) = x\) to simplify the right side of the equation. This results in: \[\text{ln}\bigg(\frac{P}{P_0}\bigg) = kt\]
04

- Solve for t

Finally, solve for \(t\) by dividing both sides of the equation by \(k\). This yields: \[t = \frac{\text{ln}\bigg(\frac{P}{P_0}\bigg)}{k}\]

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Key Concepts

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

Natural logarithm
The natural logarithm, denoted as \text{ln}, is a special logarithm with the base of the irrational number e (approximately 2.71828). It's commonly used in various mathematical and scientific contexts, especially when dealing with exponential growth or decay processes.
In the context of our problem, the natural logarithm helps in simplifying exponential equations. When we have an equation of the form \(e^{x}\), applying the natural logarithm allows us to extract the exponent by utilizing the property that \text{ln}(e^{x}) = x.
Isolating variables
Isolating variables is a crucial step in solving equations. It involves rearranging the equation so that one variable stands alone on one side. This makes it easier to solve for that variable.
In our exercise, we start with the equation \(P = P_{0} e^{kt}\). To isolate the exponential term \(e^{kt}\), we divide both sides by the initial value \(P_{0}\), resulting in \(\frac{P}{P_{0}} = e^{kt}\). This manipulation sets up the equation perfectly for the next steps.
Exponential functions
Exponential functions are mathematical functions of the form \(f(x) = a e^{bx}\), where e is the base of the natural logarithm, and a and b are constants. These functions model processes involving exponential growth or decay, such as population growth, interest calculations, and radioactive decay.
In our exercise, \(P = P_{0} e^{kt}\) is an exponential function, where \(P\) represents the final amount, \(P_{0}\) is the initial amount, \(k\) is the growth rate, and \(t\) is the time. Understanding the behavior of exponential functions is key to solving equations involving them.
Solving for a variable
Solving for a variable means finding its value by manipulating the equation. After isolating the exponential term and applying the natural logarithm, we simplify the equation to \(\text{ln}(\frac{P}{P_{0}}) = kt\).
To solve for \(t\), we need to isolate it on one side of the equation. We divide both sides by the growth rate \(k\), resulting in \(t = \frac{\text{ln}(\frac{P}{P_{0}})}{k}\). This final formula gives us the value of \(t\) in terms of \(P\), \(P_{0}\), and \(k\).
Breaking down a problem like this into manageable steps helps to understand each part fully before solving the complete equation.

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Most popular questions from this chapter

The Beer-Lambert Law. A beam of light enters a medium such as water or smoky air with initial intensity \(I_{0} .\) Its intensity is decreased depending on the thickness (or concentration) of the medium. The intensity 1 at a depth (or concentration) of \(x\) units is given by $$I=I_{0} e^{-\mu x}$$ The constant \(\mu\left(" m u^{n}\right),\) called the coefficient of absorption, varies with the medium. Use this law for Exercises 52 and \(53 .\) Light through sea water. Sea water has \(\mu=1.4\) and \(x\) is measured in meters. What would increase cloudiness more - dropping \(x\) from 2 m to 5 m or dropping \(x\) from \(7 \mathrm{m}\) to \(10 \mathrm{m} ?\) Explain.

Iodine-131 has a decay rate of \(9.6 \%\) per day. The rate of change of an amount \(N\) of iodine- 131 is given by \(\frac{d N}{d t}=-0.096 N\) where \(t\) is the number of days since the decay began. a) Let \(N_{0}\) represent the amount of iodine-131 present at \(t=0 .\) Find the exponential function that models the situation. b) Suppose that \(500 \mathrm{g}\) of iodine- 131 is present at \(t=0\) How much will remain after 4 days? c) After how many days will half of the 500 g of iodine-131 remain?

The number of women graduating from 4 -yr colleges in the United States grew from \(1930,\) when 48,869 women earned a bachelor's degree, to 2005, when approximately 832,000 women received such a degree. Find an exponential function that fits the data, and the exponential growth rate, rounded to the nearest hundredth of a percent.

The population of the United States in 1776 was about 2,508,000 In the country's bicentennial year, the population was about 216,000,000 a) Assuming an exponential model, what was the growth rate of the United States through its bicentennial year? b) Is exponential growth a reasonable assumption? Explain.

A company's total cost, in millions of dollars, is given by \(C(t)=200-40 e^{-t}\) where \(t\) is the time in years since the start-up date. (GRAPH CAN'T COPY). Find each of the following. a) The marginal \(\operatorname{cost} C^{\prime}(t)\) b) \(C^{\prime}(0)\) c) \(C^{\prime}(5) \quad\) (Round to the nearest thousand.) d) Find \(\lim _{t \rightarrow \infty} C(t)\) and \(\lim _{t \rightarrow \infty} C^{\prime}(t) .\) Why do you think the company's costs tend to level off as time passes?
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