document.write( "Question 495344: I do not know what this would be under because it is a function word problem.
\n" ); document.write( "Assuming an investment of $8,100 doubles in value every 7 years. What function rules models the worth of an investment after 28 years? What about after 35 years? \n" ); document.write( "

Algebra.Com's Answer #336068 by Theo(13342) $\"\"$ $\"About$

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f = p * 2^(n/7)
\n" ); document.write( "f = future value
\n" ); document.write( "p = present value
\n" ); document.write( "n = number of years in increments of 7\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "the formula works like this:
\n" ); document.write( "p = 8100
\n" ); document.write( "f = what you want to find.
\n" ); document.write( "n = 28
\n" ); document.write( "formula becomes:
\n" ); document.write( "f = 8100 * 2^(28/7) which becomes:
\n" ); document.write( "f = 8100 * 2^4 which becomes:
\n" ); document.write( "f = 8100 * 16 which becomes:
\n" ); document.write( "f = 129600
\n" ); document.write( "since 8100 doubles every 7 years, then you get:
\n" ); document.write( "

\r\n" );
document.write( "time point   investment\r\n" );
document.write( "    1           8100\r\n" );
document.write( "    7          16200\r\n" );
document.write( "   14          32400\r\n" );
document.write( "   21          64800\r\n" );
document.write( "   28         129600\r\n" );
document.write( "

\n" ); document.write( "the difference between each time point is equal to 7 years.
\n" ); document.write( "each time point is equivalent to the value of n in the equation.
\n" ); document.write( "time point 0 is when you invest the money
\n" ); document.write( "time point 7 is when your money doubles.
\n" ); document.write( "time point 14 is when your money doubles again.
\n" ); document.write( "time point 21 is when your money doubles again.
\n" ); document.write( "time point 28 is when your money doubles again.
\n" ); document.write( "
\n" ); document.write( "you can see that the formula will work for any value of n, except we're restricting the value of n to increments of 7 so the exponent will always be an integer.
\n" ); document.write( "we don't need to do that, however.
\n" ); document.write( "if we pick any other value of n, we can calculate the future value as well.
\n" ); document.write( "for example:
\n" ); document.write( "if n = 3, then the formula of:
\n" ); document.write( "f = 8100 * 2^(n/7) becomes:
\n" ); document.write( "f = 8100 * 2^(3/7) which becomes:
\n" ); document.write( "f = 8100 * 1.345900193 which becomes:
\n" ); document.write( "f = 10901.79156
\n" ); document.write( "in fact, we can solve for the yearly interest rate that causes our money to double every 7 years.
\n" ); document.write( "that interest rate is calculated as follows:
\n" ); document.write( "f = p * (1+x)^7
\n" ); document.write( "we let f = 2 and p = 1 to get:
\n" ); document.write( "2 = 1 * (1+x)^7 which becomes:
\n" ); document.write( "2 = (1+x)^7
\n" ); document.write( "we take the 7th root of both sides of our equation to get:
\n" ); document.write( "2^(1/7) = 1 + x
\n" ); document.write( "we subtract 1 from both sides of our equation to get:
\n" ); document.write( "2^(1/7) - 1 = x
\n" ); document.write( "we use our calculator to solve for 2^(1/7) to get:
\n" ); document.write( "1.104089514 - 1 = x
\n" ); document.write( "we simplify to get:
\n" ); document.write( "x = .104089514
\n" ); document.write( "that's our annual interest rate that will cause our money to double every 7th year.
\n" ); document.write( "to test this out, we go back to our future worth formula of:
\n" ); document.write( "f = p * (1+x)^n
\n" ); document.write( "we let p = 8100
\n" ); document.write( "we let x = .104089514
\n" ); document.write( "we let n = 28
\n" ); document.write( "we get:
\n" ); document.write( "f = 8100 * (1.104089514)^28 which becomes:
\n" ); document.write( "f = 8100 * 16 which becomes:
\n" ); document.write( "f = 129600
\n" ); document.write( "we get the same answer.
\n" ); document.write( "either of these formulas will work.
\n" ); document.write( "the more traditional formula would be the last one i gave you just now, although the first formula i gave you works fine as well.
\n" ); document.write( "notice that when n = 3, the formula becomes:
\n" ); document.write( "f = 8100 * (1.104089514)^3 which becomes:
\n" ); document.write( "f = 8100 * 1.345900193 which becomes:
\n" ); document.write( "f = 10901.79156
\n" ); document.write( "that's the same answer we got using the other formula, so either formula will get you the same answer.
\n" ); document.write( "the function you are looking for is:
\n" ); document.write( "f(n) = 8100 * (1.104089514)^n
\n" ); document.write( "n is the number of years.
\n" ); document.write( "go out 28, or 35, or any multiple of 7 and you'll see that the money doubles each time.
\n" ); document.write( "you could also use:
\n" ); document.write( "f(n) = 8100 * 2^(n/7)
\n" ); document.write( "either one will get you the same answer.
\n" ); document.write( "n is the number of years again.
\n" ); document.write( "f(n) represents the future value of the investment.
\n" ); document.write( "this is called f(n) rather than f(x) because n is the independent variable.
\n" ); document.write( "if we wanted to make it f(x), then the formula would have had to be:
\n" ); document.write( "f(x) = 8100 * 2^(x/7)\r
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