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$$\text { Solve each formula for the indicated variable.}$$ $$A=P\left(1+\frac{r}{n}\right)^{n t}, \text { for } t$$

Short Answer

Expert verified
\(t = \frac{\ln\left(\frac{A}{P}\right)}{n \cdot \ln\left(1+\frac{r}{n}\right)}\)

Step by step solution

01

Understand the Formula

We start with the formula \(A = P\left(1+\frac{r}{n}\right)^{nt}\) and we are tasked with solving for the variable \(t\). This formula is used to calculate the amount \(A\) in compound interest.
02

Isolate the Power

First, divide both sides of the equation by \(P\) to isolate the term with \(t\) in the exponent: \(\frac{A}{P} = \left(1+\frac{r}{n}\right)^{nt}\).
03

Use Logarithms

To solve for \(t\), take the natural logarithm on both sides to bring down the exponent: \(\ln\left(\frac{A}{P}\right) = nt \cdot \ln\left(1+\frac{r}{n}\right)\).
04

Solve for t

To isolate \(t\), divide both sides by \(n \cdot \ln\left(1+\frac{r}{n}\right)\): \[t = \frac{\ln\left(\frac{A}{P}\right)}{n \cdot \ln\left(1+\frac{r}{n}\right)}\].

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Key Concepts

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

Solving for a Variable
When working with equations, one often needs to isolate a particular variable. This process is called "solving for a variable." It's essential because, in many situations, you might know the other values but need to determine an unknown.
For example, in the compound interest formula, you might know the initial amount, interest rate, and final amount but need to determine the time period. Solving for a variable like this typically involves rearranging the equation. You perform operations that will "isolate" the unknown variable on one side of the equation.
Here’s a simplified approach to solving for a variable:
  • Identify the variable you want to solve for.
  • Perform inverse operations to move other terms to the opposite side of the equation.
  • Continue rearranging until the variable is isolated.
  • Ensure every step is mathematically valid to maintain equality.
With practice, solving for a variable becomes an intuitive process, helping you untangle even the most complex equations.
Logarithms in Algebra
Logarithms are a powerful mathematical concept used to solve exponential equations by using their property of turning products into sums. This is particularly helpful when dealing with equations where the variable is an exponent, as it is in our compound interest formula.
For instance, in the equation \( \ln \left( \frac{A}{P} \right) = nt \cdot \ln \left( 1+\frac{r}{n} \right) \), logarithms enable us to 'bring down' the exponent. This means transforming the equation so the unknown variable, \( t \), which is in the exponent, appears in a linear form, making it easier to solve.
  • Natural Logarithms (ln): A logarithm with the base of \( e \), where \( e \approx 2.71828 \). Used extensively in calculus and exponential growth calculations.
  • Logarithm Properties: \( \ln(ab) = \ln a + \ln b \) and \( \ln(a^b) = b \cdot \ln a \).
  • Applications: Commonly used to calculate time in interest formulas, like compounding interest, where time is the exponential argument.
Understanding logarithms unlocks the ability to solve equations that seem insurmountable at first glance.
Exponential Equations
Exponential equations are those where variables appear in the exponent, such as \( a^x = b \). These equations are prevalent in many fields including finance, physics, and biology due to the nature of exponential growth and decay.
Exponential equations often model situations where the growth rate of a value is proportional to the current value, such as in compound interest. In these situations, small changes can result in significant effects over time.
  • Compound Interest Example: The formula \( A = P(1 + \frac{r}{n})^{nt} \) represents how investments grow over discrete compounding periods.
  • Solving Techniques: Typically, these involve isolating the variable within the exponent. Often, logarithms are used to solve these equations by converting them into a simpler linear form.
  • Real-world Applications: Beyond finance, biology uses these equations to model population growth, and physics uses them in radioactive decay problems.
Mastering exponential equations is crucial for understanding and predicting growth patterns over time in various domains.

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Most popular questions from this chapter

In real life, populations of bacteria, insects, and animals do not continue to grow indefinitely. Initially, population growth may be slow. Then, as their numbers increase, so does the rate of growth. After a region has become heavily populated or saturated, the population usually levels off because of limited resources. This type of growth may be modeled by a logistic function represented by $$f(x)=\frac{c}{1+a e^{-b x}}$$ where \(a, b,\) and \(c\) are positive constants. As age increases, so does the likelihood of coronary heart disease (CHD). The fraction of people \(x\) years old with some CHD is approximated by $$f(x)=\frac{0.9}{1+271 e^{-0.122 x}}$$ (Source: Hosmer, D. and S. Lemeshow, Applied Logistic Regression, John Wiley and Sons.) (a) Evaluate \(f(25)\) and \(f(65) .\) Interpret the results. (b) At what age does this likelihood equal \(50 \% ?\)

If a line has nonzero slope \(a\), what is the slope of its reflection across the line \(y=x ?\)

Radioactive cesium 137 was emitted in large amounts in the Chernobyl nuclear power station accident in Russia on April \(26,1986 .\) The amount of cesium 137 remaining after \(x\) years in an initial sample of 100 milligrams can be described by $$A(x)=100 e^{-0.02295 x}$$ (Source: Mason, C., Biology of Freshwater Pollution, John Wiley and Sons.) (a) Estimate how much is remaining after 50 years. Is the half-life of cesium 137 greater or less than 50 years? (b) Estimate the half-life of cesium 137 .

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?

Newton's law of cooling says that the rate at which an object cools is proportional to the difference \(C\) in temperature between the object and the environment around it. The temperature \(f(t)\) of the object at time t in appropriate units after being introduced into an environment with a constant temperature \(T_{0}\) is $$f(t)=T_{0}+C e^{-k t}$$ where \(C\) and \(k\) are constants. Use this result. A piece of metal is heated to \(300^{\circ} \mathrm{C}\) and then placed in a cooling liquid at \(50^{\circ} \mathrm{C}\). After 4 minutes, the metal has cooled to \(175^{\circ} \mathrm{C}\). Estimate its temperature after 12 minutes.

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