SOLUTION: The lesson is all about extracting the roots so correct me if im in the wrong lessons... But anyway hope anyone can solve my another question :) When the amount of (p) is invest

Question 979326: The lesson is all about extracting the roots so correct me if im in the wrong lessons...
But anyway hope anyone can solve my another question :)
When the amount of (p) is invested at annual interest rate (r) compounded for (t) years, it will grow to an amount (a) given by the formula
A=P (1+r)^t
1 Suppose php 20,000 is invested at an interested rate (r) compounded annually in 5 years, it will grow to php 60,000. What is the interest rate?
2 suppose in five years, the php 35,000 invested grows to php 75 345.75 , what is the interest rate?
Found 2 solutions by rothauserc, Boreal:
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
solve for r, A=P (1+r)^t
A/P = (1+r)^t
(A/P)^(1/t) = 1+r
r = (A/P)^(1/t) - 1
*************************************************************
1) r = (60000/20000)^(1/5) - 1 = 0.24573094 approx 0.25
2) r = (75345.75/35000)^(1/5) - 1 = 0.165730456 approx 0.17

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
P=Po{!+r/n}^nt
60000=20000{1+r}^5
3= (1+r)^5
ln of both sides
1.0986=5 ln (1+r)
divide by 5 without rounding.
0.2197= ln (1+r)
Raise both to e power
1.2457= 1+ r
0.2457=r or 24.6%
1.246^5=3.003, so this checks.
second one will be less since doubling and not trebling after 5 years.
75345.75=35000 {1+r}^5
divide by 35000 and take ln of both sides
ln 2.153= 5 ln (1+r)
0.7667=5 ln (1+r)
divide by 5 and raise both to e power
1.1657= 1+r
r=0.1657 or 16.6%
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