SOLUTION: 1408=246(1-e^(-9.61*10^-7*t)) I want to solve for t, what is the procedure and what is the result?

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: 1408=246(1-e^(-9.61*10^-7*t)) I want to solve for t, what is the procedure and what is the result?      Log On


   



Question 892191: 1408=246(1-e^(-9.61*10^-7*t)) I want to solve for t, what is the procedure and what is the result?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
There is no real result, but maybe there is a typo.
1408=246(1-e^(-9.61*10^-7t)) does not look good as 1408=246%281-e%5E%28-9.61%2A10%5E-7t%29%29 ,
but I could write it as 1408=246%281-e%5E%28-0.000000961%2At%29%29
To solve, the equation could be transformed to
1408=246-246%2Ae%5E%28-0.000000961%2At%29%29-->1408-246=-246%2Ae%5E%28-0.000000961%2At%29%29-->1162=-246%2Ae%5E%28-0.000000961%2At%29%29
That cannot be true for any value of the variable because exponential functions are always positive: e%5Eanything%3E0 .
That would make the right side of the equation negative, and the left side is positive.

Curves of the type f%28t%29=A%281-e%5E%28-k%2At%29%29%29
Have f%280%29=0 and asymptotically approach A as t increases.
If the equation was 1408=2460%2A%281-e%5E%28-0.000000961%2At%29%29%29 ,
then the graph of f%28t%29=2460%281-e%5E%28-0.000000961%2At%29%29%29 would look like this: graph%28300%2C300%2C-0.2%2C0.8%2C-300%2C2700%2C2460%2A%281-e%5E%28-5x%29%29%2C2460%29
and we could solve it like this:
1408=2460-2460%2Ae%5E%28-0.000000961%2At%29

1408=2460-2460%2Ae%5E%28-0.000000961%2At%29

1408-2460=-2460%2Ae%5E%28-0.000000961%2At%29

-1052=-2460%2Ae%5E%28-0.000000961%2At%29%29

1052=2460%2Ae%5E%28-0.000000961%2At%29%29

1052%2F2460=e%5E%28-0.000000961%2At%29%29

0.42764=e%5E%28-0.000000961%2At%29%29

ln%280.42764%29=-0.000000961%2At

-.84947=-0.000000961%2At

%28-.84947%29%2F%28-0.000000961%29=t --> t=8.84%2A10%5E5