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How can you solve these simultaneous equations?
Ae^(5B) + C = 609.0
Ae^(20B) + C = 830.7
Ae^(35B) + C = 1070
Ae^(0B) + C = A + C = 554.8
I was unable to get my head around the logarithmic rules etc. The unknowns are A, B and C. e is the mathematical constant. Can someone please help me?
1 Answer
- ?Lv 41 decade agoFavourite answer
Ae^(5B) + C = 609.0 ..........(1)
Ae^(20B) + C = 830.7 ........(2)
By (1) - (2) we have,
A{e^(5B) - e^(20B)} = - 221.7 ........(3)
Ae^(35B) + C = 1070 ..........(4)
By (2) - (4) we have,
A{e^(20B) - e^(35B)} = - 239.3 ......(5)
By (5) / (3) we have,
{e^(20B) - e^(35B)} / {e^(5B) - e^(20B)} = 1.0794
e^(20B){1 - e^(15B)} /e^(5B){1 - e^(15B)} = 1.0794 ;
e^(15B) = 1.0794 ;
15B = ln(1.0794) = 0.0764 ;
B = 0.0764/15 = 0.0051 .
From (3) we have
A{e^(5B) - e^(20B)} = - 221.7 ;
A = - (221.7) / {e^0.0255 - e^0.102}
= 19.36 .
From (1) we have
Ae^(5B) + C = 609.0 ;
C = 609.00 - 19.36e^0.0255 = 589.2518 .
