Anonymous

0.5e^(0.92t)=0 => e^(0.92t)=0 => No definite solution of t. We can only write limit e^(0.92t)=0 t->-inf.

Anonymous

(0.5)e^(0.92t) = 0, ie., e^(0.92t) = 0. Taking ln of both sides implies ln[e^(0.92t)] = (0.92t) = ln[0]which is undefined.

Anonymous

The same answer as it was the last three times you asked the question.

Anonymous

0.5e^0.92t = 0 t = 0/(0.5e^0.92) t = 0 If you meant 0.5e^(0.92t) = 0, then 0.5e^(0.92t) = 0 e^(0.92t) = 0 there is no solution

Anonymous

There are no real solutions.

Anonymous

0.5e^(0.92t)=0 e^(0.92t)=0 No solution for t ∈ ℝ, as a^f(t) cannot be 0 or negative for t ∈ ℝ

Anonymous

No solution. e^x is strictly positive for all real x. So e^0.92t > 0 for all t. There is no t such that e^0.92t = 0. Solving algebraically: 0.5 e^0.92t = 0 e^0.92t = 0 0.92t = ln(0) ln(0) is not defined. The limit of ln(x) as x->0 is -infinity.