/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 Devise the exponential growth fu... [FREE SOLUTION] | 91Ó°ÊÓ

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Devise the exponential growth function that fits the given data; then answer the accompanying questions. Be sure to identify the reference point \((t=0)\) and units of time. How long will it take an initial deposit of \(\$ 1500\) to increase in value to \(\$ 2500\) in a saving account with an APY of 3.1 \(\%\) ? Assume the interest rate remains constant and no additional deposits or withdrawals are made.

Short Answer

Expert verified
Answer: It will take approximately 8.23 years for the initial deposit of $1500 to increase to $2500 in the given savings account.

Step by step solution

01

Define Variables

Let's define the given variables in the problem: Initial deposit, \(A_0 = \$1500\) Final balance, \(A(t) = \$2500\) Interest rate, \(r = 3.1\%=0.031\) We need to find the time in years, \(t\).
02

Determine the Exponential Growth Formula

The exponential growth function is given by: \(A(t) = A_0 \cdot (1 + r)^t\) Now, substitute the given values in the formula: \(2500 = 1500 \cdot (1 + 0.031)^t\)
03

Solve for the Time (t)

In this step, we will solve for \(t\). \(\frac{2500}{1500}=(1+0.031)^t\) Now, take the natural logarithm of both sides: \(\ln\left(\frac{5}{3}\right) = t \cdot \ln(1.031)\) Next, divide both sides by \(\ln(1.031)\): \(t = \frac{\ln\left(\frac{5}{3}\right)}{\ln(1.031)}\) Now, use a calculator to find the value of \(t\): \(t \approx 8.23\) years
04

Present the Answer

It will take approximately 8.23 years for an initial deposit of \( \$1500\) to increase in value to \( \$2500\) in a savings account with an APY of 3.1% and no additional deposits or withdrawals being made.

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Key Concepts

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

Exponential Equations
Exponential equations are mathematical expressions that describe situations where a quantity grows or decays at a rate proportional to its current value. This is the essence of exponential growth, or in the reverse, exponential decay.
In the example given in the original exercise, the formula for exponential growth function is used to predict the future balance of a savings account. The formula is as follows:
\[A(t) = A_0 \times (1 + r)^t\]
where:
  • \(A(t)\) is the amount of money at time \(t\).
  • \(A_0\) is the initial deposit.
  • \(r\) is the interest rate (expressed as a decimal).
  • \(t\) is the time in years.
To solve the exponential equation, you often isolate the term with the exponent \(t\), allowing you to find the 'time' it takes for the initial amount to grow to a certain value. In many cases, solving for \(t\) requires taking a logarithm of both sides, as was done in the step-by-step solution provided in the textbook.
Natural Logarithm
The natural logarithm is the logarithm to the base \(e\), where \(e\) is approximately equal to 2.71828. It's denoted as \(\ln\) and is a fundamental concept in mathematics, especially in the fields of calculus and complex analysis.
In the context of solving exponential equations, such as calculating how long it will take for money to increase from an initial deposit to a final amount in a savings account, the natural logarithm becomes a powerful tool. For instance, to isolate the exponent in the equation:
\[\frac{A(t)}{A_0} = (1 + r)^t\]
you take the natural logarithm of both sides. This results in:
\[\ln\left(\frac{A(t)}{A_0}\right) = \ln((1 + r)^t)\]
By exploiting the properties of logarithms, namely that \(\ln(a^b) = b \ln(a)\), this becomes:
\[t \cdot \ln(1 + r) = \ln\left(\frac{A(t)}{A_0}\right)\]
which can be rearranged to solve for \(t\). This step can sometimes be unintuitive to students, but understanding how to use the natural logarithm to undo exponential functions is key in various fields including finance, biology, and physics.
Compound Interest
Compound interest is the process of generating earnings on an asset's reinvested earnings. To work out the compound interest, you use the formula mentioned earlier in exponential equations. In essence, compound interest is interest on interest and will make a sum grow at a faster rate than simple interest, which is only calculated on the principal amount.
This concept is crucial for financial growth over time. For example, in the original exercise, the annual percentage yield (APY) of 3.1% indicates how much money you would earn from interest in one year, factoring in compounding. The formula for compound interest, assuming the interest is compounded once per year, can be simplified to:
\[A(t) = A_0 \times (1 + r)^t\]
This shows that the balance of the savings account after a certain number of years is a function of the initial principal (the original deposit), the interest rate, and the number of times that interest is compounded per unit \(t\), which is annually in this case.
Compound interest is the reason why it's important to start saving early, as the effects of compounding can significantly increase the value of savings over time. That is why this concept is often referred to as 'the eighth wonder of the world' by financiers, highlighting the exponential growth potential of investments over time.

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Most popular questions from this chapter

Determine whether the following statements are true and give an explanation or counterexample. a. \(\frac{d}{d x}(\sinh \ln 3)=\frac{\cosh \ln 3}{3}.\) b. \(\frac{d}{d x}(\sinh x)=\cosh x\) and \(\frac{d}{d x}(\cosh x)=-\sinh x.\) c. Differentiating the velocity equation for an ocean wave \(v=\sqrt{\frac{g \lambda}{2 \pi} \tanh \left(\frac{2 \pi d}{\lambda}\right)}\) results in the acceleration of the wave. d. \(\ln (1+\sqrt{2})=-\ln (-1+\sqrt{2}).\) e. \(\int_{0}^{1} \frac{d x}{4-x^{2}}=\frac{1}{2}\left(\operatorname{coth}^{-1} \frac{1}{2}-\cot ^{-1} 0\right).\)

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Find the area of the region bounded by the curve \(x=\frac{1}{2 y}-\sqrt{\frac{1}{4 y^{2}}-1}\) and the line \(x=1\) in the first quadrant. (Hint: Express \(y\) in terms of \(x\).)

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