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Chapter 4: Problem 19

Solve for the indicated variable. \(A=P(1+r)^{t}\) for \(t\) (used in finance)

Short Answer

Expert verified
The solution is \( t = \frac{\text{ln}\frac{A}{P}}{\text{ln}(1 + r)} \)

Step by step solution

01

Isolate the exponential term

Divide both sides of the equation by the principal amount, \(P\), to get the exponential term by itself: \[ \frac{A}{P} = (1 + r)^{t} \]
02

Apply the natural logarithm

Take the natural logarithm (ln) of both sides to remove the exponent: \[ \text{ln}\frac{A}{P} = \text{ln}((1 + r)^{t}) \]
03

Use the power rule of logarithms

Apply the power rule of logarithms, which states \( \text{ln}a^{b} = b \text{ln} a \), to the right side: \[ \text{ln}\frac{A}{P} = t \text{ln}(1+r) \]
04

Solve for the variable \(t\)

Isolate \(t\) by dividing both sides by \( \text{ln}(1 + r) \): \[ t = \frac{\text{ln}\frac{A}{P}}{\text{ln}(1 + r)} \]

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Key Concepts

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

Natural Logarithm
One of the key concepts in solving exponential equations is the use of the natural logarithm. The natural logarithm, denoted as \(\text{ln}\), is the logarithm to the base \(\text{e}\), where \(\text{e}\) is approximately equal to 2.71828. Natural logarithms are particularly useful for simplifying equations involving exponential terms.
In our exercise, after isolating the exponential term, we take the natural logarithm of both sides to help solve for the variable. This allows us to leverage properties of logarithms, such as the power rule.
Finance Applications
The equation \(A = P(1 + r)^{t}\) is often used in finance to calculate the future value of an investment. Here, \(\text{A}\) represents the amount of money accumulated after \(\text{t}\) years, \(P\) is the principal amount (initial investment), and \(r\) is the annual interest rate.
By solving for \(\text{t}\), we can determine how long it will take for an investment to grow to a specific amount, given a fixed interest rate. This can be crucial for planning and optimizing financial decisions.
Isolating Variables
Isolating the variable you're solving for is a critical step in algebra. It means rearranging the equation so that the variable of interest stands alone on one side. In this exercise, our goal is to solve for \(\text{t}\).
We start by isolating the exponential term. By dividing both sides by the principal amount \(P\), we get the exponential term \( (1 + r)^t \) by itself. This step simplifies the equation and sets the stage for applying logarithms.
Power Rule of Logarithms
The power rule of logarithms is another fundamental concept used here. It states that \( \text{ln}(a^b) = b \text{ln}(a) \). This property helps us move the exponent \(t\) out of the logarithm, allowing us to solve for the variable.
In our exercise, after applying the natural logarithm to both sides, we use the power rule to rewrite \( \text{ln}((1 + r)^t) \) as \( t \text{ln}(1 + r) \). This transformation is crucial for ultimately isolating \( t \) and solving the equation.

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Most popular questions from this chapter

Graph the following functions on the window [-3,3,1] by [-1,8,1] and comment on the behavior of the graphs near $$ \begin{array}{l} x=0 \\ \mathrm{Y}_{1}=e^{x} \\ \mathrm{Y}_{2}=1+x+\frac{x^{2}}{2} \\ \mathrm{Y}_{3}=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6} \end{array} $$

Show that \(-\ln \left(x-\sqrt{x^{2}-1}\right)=\ln \left(x+\sqrt{x^{2}-1}\right)\).

A table of data is given. a. Graph the points and from visual inspection, select the model that would best fit the data. Choose from $$\begin{array}{ll} y=m x+b \text { (linear) } & y=a b^{x} \text { (exponential) } \\ y=a+b \ln x \text { (logarithmic) } & y=\frac{c}{1+a e^{-b x}} \text { (logistic) } \end{array}$$ b. Use a graphing utility to find a function that fits the data. $$ \begin{array}{|c|c|} \hline x & y \\ \hline 3 & 2.7 \\ \hline 7 & 12.2 \\ \hline 13 & 25.7 \\ \hline 15 & 30 \\ \hline 17 & 34 \\ \hline 21 & 44.4 \\ \hline \end{array} $$

Compare the graphs of the functions. $$ Y_{1}=\ln (2 x) \quad \text { and } \quad Y_{2}=\ln 2+\ln x $$

Determine if the statement is true or false. For each false statement, provide a counterexample. For example, \(\log (x+y) \neq \log x+\log y\) because \(\log (2+8) \neq \log 2+\log 8\) (the left side is 1 and the right side is approximately 1.204 ). $$ \ln 10=\frac{1}{\log e} $$
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