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Help with a pre calculus problem?
An initial investment of $100,000 is left to grow in an account where interest is compounded continuously. At the end of 7 years there is $125,000 in the account.
Use the formula A=Pe^(rt) to find the value of r to three decimal places.
The problem is set up like 125000=100000e^(r*7)
(e is just the number preset on the calculator that equals about 2.718)
So, I have no idea how to go about doing this problem. Can someone help?
2 Answers
- 1 decade agoFavorite Answer
so first divide both sides by 100000
then apply the log e to both sides & you'll have ln 125000/100000=r*7
divide by 7
- ?Lv 45 years ago
this is the thank you to sparkling up utilising row-cut back x + y + z = 7 3x - 2y + z = 3 x + 6y + 3z = 25 a million a million a million | 7 3 -2 a million | 3 a million 6 3 | 25 R2 = R2 - 3 * R3 a million a million a million | 7 0 -20 -8 | -seventy two a million 6 3 | 25 R3 = R3 -R1 a million a million a million | 7 0 -20 -8 | -seventy two 0.5 2 | 18 R3 = R3 + a million/4 * R2 a million a million a million | 7 0 -20 -8 | -seventy two 0 0 0 | 0 R1 = R1 + a million/20 * R2 a million 0 .6 | 3.4 0 -20 -8 | -seventy two 0 0 0 | 0 R2 = -R2/20 a million 0 .6 | 3.4 0 a million .4 | 3.6 0 0 0 | 0 So the acceptable constraints are that: x + .6z = 3.4 y + .4z = 3.6 working example, suggestions comprise: x = a million a million + .6z = 3.4 .6z = 2.4 z = 4 y + .4z = 3.6 y + a million.6 = 3.6 y = 2 x = a million, y = 2, z = 4 x = 2 2 + .6z = 3.4 .6z = a million.4 z = 2.33.. y + .4(2.33..) = 3.6 y = 2.sixty six x = 2, y = 2.sixty six, z = 2.33
