Simplifying Logs

• January 9th 2010, 08:58 AM
mikewhant
Simplifying 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!

Express the following without logarithms:

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

Thanks again for any input

Mike
• January 9th 2010, 09:33 AM
TWiX
Quote:

Originally Posted by mikewhant
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!

Express the following without logarithms:

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

Thanks again for any input

Mike

Its
$
\frac{1}{2ln(Q+1)-3ln(R)+2}
$

or
$
\frac{1}{2ln(Q+1)}-3ln(R)+2
$

Using brackets is nicely to make something readable.

**Edit**:
OR
$
\frac{1}{2}ln(Q+1)-3ln(R)+2
$

when Amer solve your question according to it?
• January 9th 2010, 09:39 AM
Amer
Quote:

Originally Posted by mikewhant
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!

Express the following without logarithms:

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

Thanks again for any input

Mike

you know that

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

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

$\ln e = 1$

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

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

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

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

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

$P = \frac{(Q+1)^{\frac{1}{2}}}{R^3} \cdot e^2$
• January 9th 2010, 10:51 AM
mikewhant
Thats great thanks alot.

Take care

Mike