Question

-e^(-x)=c, c is a constant
does x^(-1)=ln(c)
and so x=1/ln(c) ?

Answer 1

basically I'm rusty on my laws of exponents with regard to log functions... :)

Answer 2

-e^(-x)=c -> e^(-x) = -c -> -x = ln(-c) -> x=-ln(-c)

Answer 3

no good, log of a negative number is undefined

Answer 4

C is a positive constant

Answer 5

c cannot be positive in the problem you gave. Note that e^(-x) = 1/e^x >0 so -e^(-x) = -(1/e^x) <0 so c is negative.

Answer 6

well my actual problem is 85/7=(-90/7)e^(-7t/2), what is t?

Answer 7

just wanted to simplify it for my post, t should have an actual positive solution

Answer 8

like you said, you cannot take the natural log of a negative number without involving complex numbers.

Answer 9

there is a real solution to this problem though...

Answer 10

you will get -17/18 = e^(-7t/2) you cannot take the ln of -17/18

Answer 11

I think there is a trick to moving coeficients of logs to their powers

Answer 12

You do not have any coefficients on yours logs. Until you take the ln of both sides, you do not have any logs. How do you know there is a real solution.

Answer 13

because this is the result of a cooling problem in which we are asked when will the temp = 65 degrees if you turn the heater on to 70, I'm fairly sure i've got the equation correct

Answer 14

see if you had 3ln(7) = ln(7^3), but that has nothing to do with what you are trying to solve here.

Answer 15

and the temp will obviously reach 65 after some length of time

Answer 16

I do not know much about physics or what not, but if you raised the heater to 70 degrees, how would the room? cool to 65 degrees. There would be an increase in temperature up to 70.

Answer 17

it starts at 40 degrees, but you're right I must have done my equation incorrectly

Answer 18

I am sorry, I couldn't be of more help, I do hope that you figure it out though.

Answer 19

the limit as t->infinity is 52.9 degrees and it should be 70 or close to it

Answer 20

thx