Is this correct

**subzero06** · March 16th 2009, 06:59 PM

is it?

i just want to make sure.

**Jhevon** · March 16th 2009, 07:27 PM

Originally Posted by subzero06

is it?

i just want to make sure.

Nope, not right. $\ln ({\color{red}-} 0.73)$ should send up a whole bunch of red flags, by the way

**Reckoner** · March 16th 2009, 07:37 PM

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.

$\ln e^{-0.73t}=-0.73t.$

**stapel** · March 17th 2009, 05:43 AM

Might it help to move the exponential to get a positive power...?

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

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

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

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

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

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

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

...and so forth.

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

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