document.write( "Question 613123: A bank pays 9 percent interest compounded annually. There are 1000 dollars in the account in January 1967. You may use a calculator.\r
\n" ); document.write( "\n" ); document.write( "a) How much money will be in the account in January 1968? dollars\r
\n" ); document.write( "\n" ); document.write( "b) How much money will be in the account in January 1969? dollars\r
\n" ); document.write( "\n" ); document.write( "c)How much money will be in the account in January 1970? dollars\r
\n" ); document.write( "\n" ); document.write( "d) How much money will be in the account in January 1977? dollars\r
\n" ); document.write( "\n" ); document.write( "e) How much money will be in the account x years after 1967? dollars\r
\n" ); document.write( "\n" ); document.write( "f) In what year will there first be one million dollars in the account? [Hint: use your answer from (e) and logs] in the year \n" ); document.write( "
Algebra.Com's Answer #385885 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
For annual compounding with an annual rate of increase (or decrease) the equation is:
\n" ); document.write( "Amount = (Original amount)*(1 + percent change (as a decimal))^(number of years since the start)
\n" ); document.write( "If the change is a decrease, use a negative decimal for the percent.

\n" ); document.write( "For you problem, with an original amount of 1000 and a 8% interest (which is an increase), your equation would be:
\n" ); document.write( "\"A+=+1000%2A%281%2B0.09%29%5Ex\"
\n" ); document.write( "where \"x\" is the number of years since 1967 (or the year - 1967) and A is the amount after x years.
\n" ); document.write( "Adding the 1 and the decimal, this simplifies to:
\n" ); document.write( "\"A+=+1000%2A%281.09%29%5Ex\"
\n" ); document.write( "This equation is the answer to part e.

\n" ); document.write( "For parts a-d, just figure out the number to use for x, year - 1967, put it in the equation and use your calculator to find A.

\n" ); document.write( "For part f you're given the A and asked to find x:
\n" ); document.write( "\"1000000+=+1000%281.09%29%5Ex\"
\n" ); document.write( "To solve this we will be using logarithms, as the problem suggests. But first we will make things a little easier if we isolate the base, 1.09, and its exponent, x. Dividing both sides of the equation by 1000 we get:
\n" ); document.write( "\"1000+=+%281.09%29%5Ex\"
\n" ); document.write( "Now we use logarithms. Any base of the logarithm can be used. But there are advanatges to choosing certain bases:
  • If you use a base for the logarithm that matches the base of the exponent, you will end up with the simplest possible expression of the exact answer.
  • If you use a base that your calculator \"knows\", base 10 or base e (aka ln), then you get an expression that can be easily converted into a decimal approximation.
To illustrate I will use both. First a base that matches:
\n" ); document.write( "Finding the base 1.09 logarithm of both sides of our equation we get:
\n" ); document.write( "\"log%281.09%2C+%281000%29%29+=+log%281.09%2C+%281.09%5Ex%29%29\"
\n" ); document.write( "Next we use a property of logarithms, \"log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29\", to move the exponent of the argument out in front. It is this very property that is the reason we use logarithms to begin with. By moving the exponent, where the variable is, out in front the variable is now \"out in th eopen\" and we can use regular algebra to solve for it.
\n" ); document.write( "Using this property on our equation we get:
\n" ); document.write( "\"log%281.09%2C+%281000%29%29+=+x%2Alog%281.09%2C+%281.09%29%29\"
\n" ); document.write( "By definition, \"log%28a%2C+%28a%29%29+=+1\" for all bases. So the log on the right sides is a 1. (This is why matching the bases get us the simplest expression.) So now we have:
\n" ); document.write( "\"log%281.09%2C+%281000%29%29+=+x\"
\n" ); document.write( "This is the simplest, exact expression for the solution to part f.

\n" ); document.write( "Now I'll repeat this with base e logs. Since the steps are the same, I'll leave out the commentary.
\n" ); document.write( "\"1000+=+%281.09%29%5Ex\"
\n" ); document.write( "\"ln%281000%29+=+ln%281.09%5Ex%29\"
\n" ); document.write( "\"ln%281000%29+=+x%2Aln%281.09%29\"
\n" ); document.write( "This time, the log on the right is not a 1. So we divide both sides by it:
\n" ); document.write( "\"ln%281000%29%2Fln%281.09%29+=+x\"
\n" ); document.write( "This is another exact (but not so simple) expression for the solution to part f. This one, however, can be easily entered into your calculator to find a decimal approximation. I'll leave it up to you and your calculator and the following notes:
  • Do NOT divide 1000 by 1.09 first and then find the log. Find the ln(1000) and the ln(1.09) first and then divide the logs. If you have a calculator that has buttons for parentheses, then you could just type in the whole expression at once:
    \n" ); document.write( "ln(1000)/ln(1.09)
  • The answer you get will be a decimal. Since the compounding is just once a year, round this number up to the next whole number if the decimal is anything but a zero! This number is NOT the answer. It is just the number of years after 1967. The problem asks for the year. So add your rounded number to 1967 to get the year number.
    \n" ); document.write( "Let's pretend your calculator came up with 15.03442 for x. (15.03442 is NOT the actual number you should get from your calculator. It is just being used as an example. The actual number is much larger.) Round this up to 16. Then the year when there would be at least 1 million dollars would be: 1967+16 = 1983. (Since the decimal part of x was so small, 0.03442, it means that in 1982, there would be almost but not quite a million dollars. This is why we round up any decimal part (except 0).
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