System of three equations with $e$ I'm trying to solve the following system of equations:
$0.001213=A+B.e^{30C}$
$0.002817=A+B.e^{40C}$
$0.00793=A+B.e^{50C}$
I tried elimination method but I got no answer. What I did was multiplying the first equation with -1 and summing it.
Could you please help me on how to solve this?
Thanks

Pay attention the pattern of $e^{30C},e^{40C},e^{50C}$ and notice 40-30=50-40, which gives us a hint  where to start.
$0.001213=A+B.e^{30C}$(I)
$0.002817=A+B.e^{40C}$ (II)
$0.00793=A+B.e^{50C}$  (III)
(II)-(I) $0.001604=B.(e^{40C}-e^{30C})$ =$B.e^{30C}(e^{10C}-1)$ (IV)
(III)-(II) $0.005133=B.(e^{50C}-e^{40C})$=$B.e^{40C}(e^{10C}-1)$ (V)
(V)/(IV) $\frac{5133}{1604}$=$\frac{e^{40C}}{e^{30C}}$=$e^{10C}$, $ln \frac{5133}{1604}$=10C $\to$
C= $\frac{ln \frac{5133}{1604}}{10}$
With C known, substract (I) and (II) to find B. With B,C known, use (I) to find A.
