Hi there, I am a bit stuck as to how to do this question. I know the laws of logs but this has me stumped!

Thanks in advance

Express the following without logarithms:

lnP=1/2ln(Q+1)-3lnR+2

Thanks again for any input

Mike

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- Jan 9th 2010, 08:58 AMmikewhantSimplifying Logs
Hi there, I am a bit stuck as to how to do this question. I know the laws of logs but this has me stumped!

Thanks in advance

Express the following without logarithms:

lnP=1/2ln(Q+1)-3lnR+2

Thanks again for any input

Mike - Jan 9th 2010, 09:33 AMTWiX
- Jan 9th 2010, 09:39 AMAmer
you know that

$\displaystyle \log a + \log b = \log ab $

$\displaystyle \log a - \log b = \log \frac{a}{b} $

$\displaystyle \ln e = 1 $

$\displaystyle a\log b = \log b^a $

$\displaystyle \ln P = 1/2 \ln (Q+1) - 3\ln R + 2 $

$\displaystyle \ln P = \ln (Q+1)^{\frac{1}{2}} - \ln R^3 + 2 \ln e $

$\displaystyle \ln P = \ln \frac{(Q+1)^{\frac{1}{2}}}{R^3} + \ln e^2 $

$\displaystyle \ln P = \ln \left(\frac{(Q+1)^{\frac{1}{2}}}{R^3} \cdot e^2 \right) $

$\displaystyle P = \frac{(Q+1)^{\frac{1}{2}}}{R^3} \cdot e^2 $ - Jan 9th 2010, 10:51 AMmikewhant
Thats great thanks alot.

Take care

Mike