Solve for n over the real numbers:
4000 *1.015^n = 3542* 1.04^n
4000 1.015^n = 2^(5 - 3 n) 5^(3 - 2 n) 203^n and 3542 1.04^n = 1771 (13/25)^n 2^(n + 1):
2^(5 - 3 n) 5^(3 - 2 n) 203^n = 1771 (13/25)^n 2^(n + 1)
Take the natural logarithm of both sides and use the identities log(a b) = log(a) + log(b) and log(a^b) = b log(a):
log(2) (5 - 3 n) + log(5) (3 - 2 n) + log(203) n = -(log(25/13) n) + log(2) (n + 1) + log(1771)
Expand and collect in terms of n:
(-3 log(2) - 2 log(5) + log(203)) n + 5 log(2) + 3 log(5) = -(log(25/13) n) + log(2) (n + 1) + log(1771)
Expand and collect in terms of n:
(-3 log(2) - 2 log(5) + log(203)) n + 5 log(2) + 3 log(5) = (log(2) - log(25/13)) n + log(2) + log(1771)
Subtract n (log(2) - log(25/13)) + 5 log(2) + 3 log(5) from both sides:
(log(25/13) - 4 log(2) - 2 log(5) + log(203)) n = -4 log(2) - 3 log(5) + log(1771)
Divide both sides by log(25/13) - 4 log(2) - 2 log(5) + log(203):
Answer: |n = (-4 log(2) - 3 log(5) + log(1771))/(log(25/13) - 4 log(2) - 2 log(5) + log(203))
=4.9976294057298984......
+9479 \(1.015^{n}(4000) = 1.04^{n}(3542)\)
Divide both sides by 4000.
\(1.015^{n} = 1.04^{n}(\frac{3542}{4000})\)
Divide both sides by 1.04^n
\(\frac{1.015^{n}}{1.04^{n}}=\frac{3542}{4000}\)
That is:
\((\frac{1.015}{1.04})^{n}=\frac{3542}{4000}\)
\((\frac{1.015}{1.04})^{n}=0.8855\)
n equals the log base (1.015/1.04) of 0.8855, but since most calculators can only do base 10 or base e, we will need to use a change of base formula.
This page talks about the change of base formula: http://www.purplemath.com/modules/logrules5.htm
That tells you that:
n = [ln 0.8855]/[ln (1.015/1.04)]
Just plug that into a calculator to get that
n ≈ 4.998
+129852 It appears that the "Guest" took the "long way home".....LOL!!!!!
1.015^n * 4000 = 1.04^n * 3542 rearrange as
1.015^n / 1.04^n = 3542 / 4000
[1.015/1.04]^n = 3542 / 4000 log both sides
log [1.015/1.04]^n = log [ 3542 / 4000 ]
n * log [1.015/1.04]= log [ 3542 / 4000 ]
n = log [ 3542 / 4000 ] / log [1.015/1.04] ≈ 4.99763
