Having trouble with a section in my summer assaignment, and my teacher doesnt respond to his e-mails... If someone could help with these few questions, it would help a lot. Thanks.
Solve for y in terms of t:
1)ln(50-3y)=-.5t+ln5
2)2e^(-y/3)=t+4
3)(t/4)-ln3=ln(20-y)
4)ln(10-y)= -2t+ln8
There is also another problem I have but am not sure if I did it correctly
(16-x^2)/(x^2+x-6)>or equal to 0
In the en I got x > or equal to 22, but I have no clue if what I did was the right method to use.
Hmmm. Are you paying for the course?
the one you did:
(4-x)(4+x)/(x+3)(x-2) > 0
Totally wrong.
which means the left side is positive, which means
-all terms are postive
-two or four terms are negative.
for all positive, than means x<4 and x>-4 (see numerator), and x>-3 and x>2 see denominator. For this all to be true, then 4>x>2 this is one part of the solution.
Now, if two are negative, look at the numerator: x>4 or x<-4, and in the denominator, x<-3 OR x<2
which means, x<-4 or x>4
check that
Now for four negatives, well you can do that one.
4) ln(10-y)-ln8 = 2t
ln((10-y)/8) =2t
take the antilog of each side
e^2t=(10-y)/8
solve for y.