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Mathematics 8 Online
OpenStudy (anonymous):

i'll type my question in the comments :D

OpenStudy (anonymous):

\[\frac{ T }{ T_{0} }=(\frac{ P }{ P_{0} })^\frac{ k-1 }{ k }\] solve for k sooo i did log P/Psub 0 on both sides... i got... k=\[k=\frac{ 1 }{ 1-\log_{ \frac{ P }{ P_0 }} \frac{ T }{ T_0 }}\] is that correct?

OpenStudy (anonymous):

the answer according to the book is ln(P/P_0) over ln(PT_0/TP_0)

OpenStudy (anonymous):

are they equivalent?

OpenStudy (anonymous):

@Coolsector ? :D

OpenStudy (anonymous):

@Hero ? :D

OpenStudy (anonymous):

yes that is fine as well

OpenStudy (anonymous):

really? yay! thank you!

OpenStudy (anonymous):

your book: \[\frac{ \ln (p/p0) }{ \ln(pT0/p0T) } = \frac{ \ln (p/p0) }{ \ln(p/p0) - \ln(T/T0)) }\] now you multiply num and denom by 1/ln(p/p0) and get:

OpenStudy (anonymous):

\[\frac{ 1 }{ 1 - \ln(T/T0)/\ln(p/p0) } = \frac{ 1 }{ 1 - Log _{p/p0}(T/T0) } \]

OpenStudy (anonymous):

which is what you have :)

OpenStudy (anonymous):

yay that made it really clear! thank you!!

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