College Algebra Colorado State University

At the beginning of the chapter we were given India’s population of 1.14 billion in the year 2008 and a percent growth rate of 1.34%. Using 2008 as our starting time (t = 0), our initial population will be 1.14 billion. Since the percent growth rate was 1.34%, our value for \(r\) is 0.0134. Using the basic formula for exponential growth \(f(x)=a(1+r)^{x}\) we can write the formula, \(f(t)=1.14(1+0.0134)^{t}\text{.}\)

To estimate the population in 2020, we evaluate the function at \(t = 12\text{,}\) since 2020 is 12 years after 2008. \(f(12)=1.14(1+0.0134)^{12}\approx 1.337\) billion people in 2020

For comparison, the actual population of India in 2020 was about 1.380 billion.

First, we must notice that the interest rate is an annual rate, but is compounded monthly, meaning interest is calculated and added to the account monthly. To find the monthly interest rate, we divide the annual rate of 1.2%by 12 since there are 12 months in a year: \(\dfrac{1.2\%}{12}= 0.1\%\text{.}\) Each month we will earn 0.1% interest. From this, we can set up an exponential function, with our initial amount of $1000 and a growth rate of \(r = 0.001\text{,}\) and our input \(m\) measured in months. \(f(m)=1000(1+\dfrac{.012}{12})^{m}\) or \(f(m)=1000(1+0.001)^{m}\text{.}\) So, after 24 months, the account will have grown to \(f(24)=1000(1+0.001)^{24}=\$1024.28\text{.}\)

With radioactive decay, instead of the quantity increasing at a percent rate, the quantity is decreasing at a percent rate. Our initial quantity is \(a = 100 \text{ mg}\text{,}\) and our growth rate will be negative 13%, since we are decreasing: \(r = -0.13\text{.}\) This gives the equation: \(Q(d)=100(1-0.13)^{d}=100(0.87)^{d}\) This can also be explained by recognizing that if 13% decays, then 87% remains.

After one week, 7 days, the quantity remaining would be \(Q(7)=100(0.87)^{7}=37.73\text{,}\) so 37.73 mg of Bismuth-210 remains.

Interpreting this from the basic exponential form, we know that 86 is our initial value. This means that on January 1, 2016 this region had 86,000 Android smart phone contracts. Since \(b = 1 + r = 1.64\text{,}\) we know that every quarter the number of smart phone contracts grows by 64%. \(T(2) = 231.3056\) means that in the 2nd quarter (or at the end of the second quarter) there were approximately 231,306 Android smart phone contracts.

By defining our input variable to be \(t\text{,}\) years after 2009, the information listed can be written as two input-output pairs: \((0,80)\) and \((6,180)\text{.}\) Notice that by choosing our input variable to be measured as years after 2009, we have effectively "given" ourselves the initial value for the function: \(a = 80\text{.}\) This gives us an equation of the form \(f(t)=80b^{t}\text{.}\)

Substituting in our second input-output pair allows us to solve for \(b\text{:}\) \(180=80b^{6}\text{.}\) Dividing both sides by by 80 gives us \(b^{6}=\dfrac{180}{80}=\dfrac{9}{4}\text{.}\) Taking the 6th root of both sides gives us \(b=(\dfrac{9}{4})^{\frac{1}{6}}\approx1.1447\text{.}\)

This gives us our equation for the population: \(f(t)=80(1.1447)^{t}\text{.}\)

Recall that since \(b = 1+r\text{,}\) we can interpret this to mean that the population growth rate is \(r = 0.1447\text{,}\) and so the population is growing by about 14.47% each year.

Since we don’t have the initial value, we will take a general approach that will work for any function form with unknown parameters: we will substitute in both given input-output pairs in the function form \(f(x)=ab^{x}\) and solve for the unknown values, \(a\) and \(b\text{.}\) Substituting in \((-2, 6)\) gives \(6=ab^{-2}\text{,}\) and substituting in \((2, 1)\) gives \(1=ab^{2}\text{.}\)

We now solve these as a system of equations. To do so, we could try a substitution approach, solving one equation for a variable, then substituting that expression into the second equation. Solving \(6=ab^{-2}\) for \(a\text{:}\) \(a=\dfrac{6}{b^{-2}}=6b^{2}\text{.}\)

In the second equation, \(1=ab^{2}\text{,}\) we substitute the expression above for \(a\text{:}\) \(1=(6b^{2})b^{2}=6b^{4}\text{,}\) so \(b^{4}=\dfrac{1}{6}\text{,}\) meaning \(b=\dfrac{1}{6}^{\frac{1}{4}}\approx 0.6389\text{.}\)

Going back to the equation \(a=6b^{2}\) lets us find \(a\text{:}\) \(a=6b^{2}=6(0.6389)^{2}=2.4492\)

Putting this together gives the equation \(f(x)=2.4492(0.6389)^{x}\text{.}\)

The initial value for the function is not clear in this graph, so we will instead work using two clearer points. There are three clear points: \((-1, 1)\text{,}\) \((1, 2)\text{,}\) and \((3, 4)\text{.}\) As we saw in the last example, two points are sufficient to find the equation for a standard exponential, so we will use the latter two points.

Substituting in \((1,2)\) gives \(2=ab^{1}\text{,}\) and substituting in \((3,4)\) gives \(4=ab^{3}\text{.}\)

Solving the first equation for a gives \(a=\dfrac{2}{b}\text{,}\) and substituting this expression for \(a\) into the second equation gives: \(4=\dfrac{2}{b}b^{3}=2b^{2}.\) Solving this equation for \(b\) gives us \(b=\pm \sqrt{2}\text{,}\) but since we restrict \(b\) to only take on positive values, we have \(b=\sqrt{2}\text{.}\) We can then go back and find \(a\text{:}\) \(a=\dfrac{2}{b}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\text{.}\)

This gives us a final equation of \(f(x)=\sqrt{2}(\sqrt{2})^{x}\text{.}\)

Since we are starting with $3000, \(a = 3000\text{.}\) Our interest rate is 3%, so \(r = 0.03\text{.}\) Since we are compounding quarterly, we are compounding 4 times per year, so \(k = 4\text{.}\) We want to know the value of the account in 10 years, so we are looking for \(A(10)\text{,}\) the value when \(t = 10\text{.}\)

\(A(10)=3000(1+0.03/4)^{4(10)}=\$4045.05\text{.}\)

The account will be worth $4045.05 in 10 years.

Since the account is earning 6%, \(r = 0.06\text{.}\) Since interest is compounded twice a year, \(k = 2\text{.}\)

In this problem, we don’t know how much we are starting with, so we will be solving for \(a\text{,}\) the initial amount needed. We do know we want the end amount to be $40,000, so we will be looking for the value of a so that \(A(18) = 40,000\text{.}\) \(40,000=A(18)=a(1+\dfrac{0.06}{2})^{2(18)}\text{,}\) meaning that \(40,000=a(2.8983)\text{.}\) Solving this for \(a\) gives \(a=\dfrac{40,000}{2.8983}\approx \$13,801\text{.}\)

So, Lily will need to invest $13,801 to have $40,000 in 18 years.

We can compare these rates using the annual percentage yield - the actual percent increase in a year. Bank A: \(\text{APY}=(1+\dfrac{0.012}{4})^{4}-1=0.012054 = 1.2054\%\) Bank B: \(\text{APY}=(1+\dfrac{0.011}{12})^{12}-1=0.011056 = 1.1056\%\)

Bank B’s monthly compounding is not enough to catch up with Bank A’s better APR, so Bank A offers a better rate.

Since we are given a continuous decay rate, we use the continuous growth formula. Since the substance is decaying, we know the growth rate will be negative: \(r = -0.173\text{.}\) \(f(3)=100e^{-0.173(3)}\approx 59.512\text{.}\) So 59.512 mg of Radon-222 will remain.

Here, the continuous growth rate is 10%, so \(r = 0.10\text{.}\) We start with $1000, so \(a = 1000\text{.}\) To find the value after 1 year, \(f(1)=1000e^{0.10(1)}\approx \$1105.17\text{.}\)

Notice this is a \(\$105.17\) increase for the year. As a percent increase, this is \(\dfrac{105.17}{1000}=0.10517=10.517\%\) increase over the original $1000.