Question 152528
<u><b>Part A:</b></u> We can just plug in the numbers:
The principal amount (P) is 3000, since that is what was originally deposited. 
The rate (r) is 0.06 because 6% means 6/100. 
The number (n) that it is compounded is 1, since annually means only once a year.
The time (t) is 9 years. 
<br>The formula is much easier to deal with when we first plug in n=1, so let's start with that:
{{{A = P(1+(r/n))^nt}}}
{{{A = P(1+(r/1))^(1t)}}}
{{{A = P(1+r)^t}}}
<br>This is also the general formula for the return on any deposit compounded annually. Now we can plug in the specific numbers:
{{{A = 3000(1+0.06)^9}}}
{{{A = 3000(1.06^9)}}}
{{{A = 3000(1.689478)}}}
{{{A = 5068.436877}}}
<br>The return is $5068.44.
<br><br><u><b>Part B:</b></u> We can just plug in the numbers. P, r, and t are the same, but now n changes from 1 to 4:
{{{A = P(1+(r/n))^nt}}}
{{{A = P(1+(r/4))^(4t)}}}
<br>This is also the general formula for the return on any deposit compounded quarterly. Now let's plug in the numbers:
{{{A = 3000(1+(0.06/4))^(4*9)}}}
{{{A = 3000(1+0.015)^36}}}
{{{A = 3000(1.015)^36}}}
{{{A = 3000(1.709139)}}}
{{{A = 5127.418614}}}
<br>The return is $5127.42
<br><br><u><b>Part C:</b></u> Compounding quarterly yields more interest. This is because when we do it once a year, it only multiplies the whole thing once by 1.06. When we do it four times a year, it multiplies it by 1.015^4, which is 1.06136355, which is actually more than 1.06.
<br><br><u><b>Part D:</b></u> Now we will use a different formula entirely.
{{{A = Pe^rt}}}
<br>We can still plug in the same numbers for P, r, and t.
{{{A = 3000e^(9*0.06)}}}
{{{A = 3000e^(.54)}}}
{{{A = 3000(1.7160069)}}}
{{{A = 5148.020586}}}
<br>The return is $5148.02
<br><br><u><b>Part E:</b></u> In order to find out how much it will take to double the money, we start with the equation: 
{{{A = 2P = Pe^rt}}}
<br>The variables P and r will be the same, but we no longer know how much time it will take.
{{{2(3000) = 3000e^(0.06*t)}}}
{{{6000 = 3000e^(0.06*t)}}}
{{{6000/3000 = (3000e^(0.06*t))/3000}}}
{{{2 = e^(0.06*t)}}}
<br>It looks like we need to use logarithms to solve this problem. The natural log of 2 will equal 0.06*t:
{{{ln (2) = 0.06*t}}}
{{{(ln (2))/0.06 = t}}}
{{{t = 11.552453}}}
<br> It should take 11.55 years for the deposit to double.