document.write( "Question 152528This question is from textbook
\n" ); document.write( ": Please help not really understanding this problem: The formula for calculating the amount of money returned for an initial deposit into a bank account or CD is given by A=P(1+r/n)^nt
\n" ); document.write( "A is the amount of the return.
\n" ); document.write( "P is the principal amount initially deposited.
\n" ); document.write( "r is the annual interest rate (expressed as a decimal).
\n" ); document.write( "n is the number of compound periods in one year.
\n" ); document.write( "t is the number of years.
\n" ); document.write( "Carry all calculations to six decimals on each intermediate step, then round the final answer to the nearest cent.
\n" ); document.write( "Suppose you deposit $3000 for 9 years at a rate of 6%
\n" ); document.write( "a)Calculate the return(A) if the bank compounds annually (n=1). Round your answer to the hundredth's place.
\n" ); document.write( "b)Calculate the return (A) if the bank compounds quarterly (n=4). Round your answer to the hundredth's place.
\n" ); document.write( "c) Does compounding annually or quarterly yield more interest? Why?
\n" ); document.write( "d)If a bank compounds continuously, then the formula userdf is A=Pe^rt where e is a constant and equals approximately 2.7183. Calculate A with continuous compounding. Round your answer to the hundredth's place.
\n" ); document.write( "e) How long will it take to double my money? At 6% interest rate and continuous compounding , what is the answer? Round your answer to the hundredth's place.
\n" ); document.write( "Thank you so much for your help in advance. \n" ); document.write( "
Algebra.Com's Answer #112138 by mducky2(62)\"\" \"About 
You can put this solution on YOUR website!
Part A: We can just plug in the numbers:
\n" ); document.write( "The principal amount (P) is 3000, since that is what was originally deposited.
\n" ); document.write( "The rate (r) is 0.06 because 6% means 6/100.
\n" ); document.write( "The number (n) that it is compounded is 1, since annually means only once a year.
\n" ); document.write( "The time (t) is 9 years.
\n" ); document.write( "
The formula is much easier to deal with when we first plug in n=1, so let's start with that:
\n" ); document.write( "\"A+=+P%281%2B%28r%2Fn%29%29%5Ent\"
\n" ); document.write( "\"A+=+P%281%2B%28r%2F1%29%29%5E%281t%29\"
\n" ); document.write( "\"A+=+P%281%2Br%29%5Et\"
\n" ); document.write( "
This is also the general formula for the return on any deposit compounded annually. Now we can plug in the specific numbers:
\n" ); document.write( "\"A+=+3000%281%2B0.06%29%5E9\"
\n" ); document.write( "\"A+=+3000%281.06%5E9%29\"
\n" ); document.write( "\"A+=+3000%281.689478%29\"
\n" ); document.write( "\"A+=+5068.436877\"
\n" ); document.write( "
The return is $5068.44.
\n" ); document.write( "

Part B: We can just plug in the numbers. P, r, and t are the same, but now n changes from 1 to 4:
\n" ); document.write( "\"A+=+P%281%2B%28r%2Fn%29%29%5Ent\"
\n" ); document.write( "\"A+=+P%281%2B%28r%2F4%29%29%5E%284t%29\"
\n" ); document.write( "
This is also the general formula for the return on any deposit compounded quarterly. Now let's plug in the numbers:
\n" ); document.write( "\"A+=+3000%281%2B%280.06%2F4%29%29%5E%284%2A9%29\"
\n" ); document.write( "\"A+=+3000%281%2B0.015%29%5E36\"
\n" ); document.write( "\"A+=+3000%281.015%29%5E36\"
\n" ); document.write( "\"A+=+3000%281.709139%29\"
\n" ); document.write( "\"A+=+5127.418614\"
\n" ); document.write( "
The return is $5127.42
\n" ); document.write( "

Part C: 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.
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Part D: Now we will use a different formula entirely.
\n" ); document.write( "\"A+=+Pe%5Ert\"
\n" ); document.write( "
We can still plug in the same numbers for P, r, and t.
\n" ); document.write( "\"A+=+3000e%5E%289%2A0.06%29\"
\n" ); document.write( "\"A+=+3000e%5E%28.54%29\"
\n" ); document.write( "\"A+=+3000%281.7160069%29\"
\n" ); document.write( "\"A+=+5148.020586\"
\n" ); document.write( "
The return is $5148.02
\n" ); document.write( "

Part E: In order to find out how much it will take to double the money, we start with the equation:
\n" ); document.write( "\"A+=+2P+=+Pe%5Ert\"
\n" ); document.write( "
The variables P and r will be the same, but we no longer know how much time it will take.
\n" ); document.write( "\"2%283000%29+=+3000e%5E%280.06%2At%29\"
\n" ); document.write( "\"6000+=+3000e%5E%280.06%2At%29\"
\n" ); document.write( "\"6000%2F3000+=+%283000e%5E%280.06%2At%29%29%2F3000\"
\n" ); document.write( "\"2+=+e%5E%280.06%2At%29\"
\n" ); document.write( "
It looks like we need to use logarithms to solve this problem. The natural log of 2 will equal 0.06*t:
\n" ); document.write( "\"ln+%282%29+=+0.06%2At\"
\n" ); document.write( "\"%28ln+%282%29%29%2F0.06+=+t\"
\n" ); document.write( "\"t+=+11.552453\"
\n" ); document.write( "
It should take 11.55 years for the deposit to double.\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
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