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Solve the equation for n. 1.015^n * 4 000 = 1.04^n * 3 542

Guest Feb 12, 2017

Best Answer 

 #4
avatar
+5

 hectictar: Very good job!.

At the last step: (1.015/1.04)^n = 0.8855. Just take the logs of both sides to ANY base, including 10:

Thus: n x Log(1.015/1.04) = Log(0.8855) (base 10)

n =Log(0.8855) / Log(1.015/1.04)

n = -0.052811... / -0.010567297....

n =4.9976.....

Guest Feb 12, 2017
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4+0 Answers

 #1
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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......

Guest Feb 12, 2017
 #2
avatar+5931 
+6

\(1.015^{n}(4000) = 1.04^{n}(3542)\)

 

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

hectictar  Feb 12, 2017
 #3
avatar+81032 
+5

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

 

 

 

cool cool cool

CPhill  Feb 12, 2017
 #4
avatar
+5
Best Answer

 hectictar: Very good job!.

At the last step: (1.015/1.04)^n = 0.8855. Just take the logs of both sides to ANY base, including 10:

Thus: n x Log(1.015/1.04) = Log(0.8855) (base 10)

n =Log(0.8855) / Log(1.015/1.04)

n = -0.052811... / -0.010567297....

n =4.9976.....

Guest Feb 12, 2017

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