# Is this correct

• Mar 16th 2009, 06:59 PM
subzero06
Is this correct
is it?

i just want to make sure.

http://i43.tinypic.com/168tv1u.png
• Mar 16th 2009, 07:27 PM
Jhevon
Quote:

Originally Posted by subzero06
is it?

i just want to make sure.

http://i43.tinypic.com/168tv1u.png

Nope, not right. $\displaystyle \ln ({\color{red}-} 0.73)$ should send up a whole bunch of red flags, by the way
• Mar 16th 2009, 07:37 PM
Reckoner
Quote:

Originally Posted by subzero06
is it?

i just want to make sure.

Jhevon gives good advice: when possible, make sure that your answer makes sense within the context of the problem.

$\displaystyle \ln e^{-0.73t}=-0.73t.$
• Mar 17th 2009, 05:43 AM
stapel
Might it help to move the exponential to get a positive power...?

. . . . .$\displaystyle P\, +\, 56.5Pe^{-0.73t}\, =\, 230$

. . . . .$\displaystyle P\, +\, \frac{56.5P}{e^{0.73t}}\, =\, 230$

. . . . .$\displaystyle e^{0.73t}P\, +\, 56.5P\, =\, 230 e^{0.73t}$

. . . . .$\displaystyle 56.5P\, =\, 230 e^{0.73t}\, -\, e^{0.73t}P$

. . . . .$\displaystyle 56.5P\, =\, \left(230\, -\, P\right) e^{0.73t}$

. . . . .$\displaystyle \frac{56.5P}{230\, -\, P}\, =\, e^{0.73t}$

. . . . .$\displaystyle \ln{\left(\frac{56.5P}{230\, -\, P}\right)}\, =\, 0.73t$

...and so forth.

:D

P.S. To review why "ln(e^(some power))" simplifies to just the (some power), try here.