SOLUTION: Can someone help solve the below equations: Q1 2950=2000(1+r)^10 Q2 2p = p(1.09)^n Thanks.

Question 726649: Can someone help solve the below equations:
Q1
2950=2000(1+r)^10
Q2
2p = p(1.09)^n
Thanks.

Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
Q1: Given
(1) 2950 = 2000*(1+r)^10 or
(2) (1+r)^10 = 2950/2000 or
(3) (1+r)^10 = 1.475
Now take the 1/10th root of each side and get
(4) 1+r = (1.475)^(0.1)
Use your calculator to find
(5) 1+r = 1.03963... or
(6) r = 1.03963... -1 or
(7) r = 0.03963... or
(8) r = 100*(0.03963...) or
(9) r = 3.963095.. %
To check this value for r, put it into (1).
Is (2950 = 2000*(1+0.0396...)^10)?
Is (2950 = 2000*(1.0396...)^10)?
Is (2950 = 2000*(1.475))?
Is (2950 = 2950)? Yes
Q1 Answer: r = 3.9631%
Q2: Given
(10) 2p = p*(1.09)^n or assuming p is not equal to zero we get
(11) 2 = (1.09)^n
Now take the ln (or LOG) of both sides of (11) to get
(12) ln(2) = n*ln(1.09)
Solve (12) for n to get
(13) n = ln(2)/ln(1.09)
Use your calculator to evaluate (13) to get
(14) n = (0.693...)/(0.08617...) or
(15) n = 8.04323...
Check this value of n with (11)
Is (2 = (1.09)^(8.04323...))?
Is (2 = 2)? Yes
Q2 Answer: n = 8.04323...

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