Question

this one

Answer 1

http://screencast.com/t/Bj55jV3l4c
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Answer 2

\[-8pe^{\frac{-p}{2}} -16e^{\frac{-p}{2}}+16 =9\]
i came this far but i donno how to write it in dat form?

Answer 3

kind of confused with taking ln :/

Answer 4

I think we can start by solving for the e^(-p/2), since the form of the item we are trying to find is by taking a logarithm.

Answer 5

So we can factor that e^(-p/2) out from the expression:
(-8p - 16)e^(-p/2) + 16 = 9

Answer 6

cool :)

Answer 7

one more question ?
\[\ln(-8p^{\frac{-p}{2}})\]

whats the asnwer

Answer 8

Solving for e^(-p/2):
(-8p - 16) e^(-p/2) = 9-16 = -7
e^(-p/2) = -7/(-8p -16) = 7/(8p + 16) We can take logarithm of both sides here.
-p/2 = ln (7 / (8p + 16))
p = -2 ln (7 / (8p + 16) )

We need just one logarithm property here to get the form asked of us, and that is to take the -1 out front into the logarithm as a power, and flip the fraction:
\( b \ln a = \ln a^{b} \)
\( - \ln a = \ln a^{-1} = \ln \dfrac{1}{a} \)

Answer 9

oh yeah thanks btw wt bt the answer to my question?

Answer 10

I wasn't sure where ln(-8p^(-p/2)) was coming from...
but considering p was called a positive constant, -8p is a negative number which logarithms aren't defined for.

Answer 11

oh okay so if it is +p then?

Answer 12

ln(8p^(-p/2))
using logarithm properties you could break this up: ln(ab) = ln a + ln b
ln(8p^(-p/2)) = ln 8 + ln p^(-p/2)
and then we have logarithm power property:
= ln 8 - p/2 ln p
Was that what you were looking for?

Answer 13

oh ok yeah got it :)

Answer 14

Thanks man! Really grateful for your help, would have taken so long to figure out all dis doubts :)

Answer 15

Always glad to be helpful! :D