# Mass Bias Calculation

• Jan 6th 2010, 04:30 PM
Ashley22
Mass Bias Calculation
Hi I need some help just reorganizing this equation for mass bias:

Robs = Rtrue(1+e)^m

I need the equation rearranged for "e" the mass bias factor. It seems to get tricky with logarithms and I am not sure if I have it correct or not

Any help would be much appreciated
• Jan 6th 2010, 04:49 PM
TKHunny
Well, how can we tell you if you are right if you don't present anything.

Logarithms are not tricky. Stop thinking that.

$\displaystyle R_{obs} = R_{true}(1+e)^{m}$

$\displaystyle log(R_{obs}) = log(R_{true}(1+e)^{m})$

$\displaystyle log(R_{obs}) = log(R_{true}) + log((1+e)^{m})$

$\displaystyle log(R_{obs}) = log(R_{true}) + m \cdot log(1+e)$

$\displaystyle log(R_{obs}) - log(R_{true}) = m \cdot log(1+e)$

$\displaystyle log(R_{obs}/R_{true}) = m \cdot log(1+e)$

$\displaystyle \frac{1}{m}log\left(\frac{R_{obs}}{R_{true}}\right ) = log(1+e)$

You're not going to make me do ALL the work, are you?
• Jan 7th 2010, 01:44 PM
Ashley22
My apologies, whether you believe me or not I actually got to that part of the equation. What I can't figure out is how to get rid of the logarithms now. I'll show you what I have as my temporary solution:

Robs = Rtrue(1+e)^m
Robs/Rtrue = (1+e)^m
log (Robs/Rtrue) = log(1+e)^m
log (Robs/Rtrue) = (m)log(1+e)
log (Robs/Rtrue)/m = log(1+e)

Now this is where my confidence ends, to get the "e" out of the logarithm, I believe you need to put the logarithm to the base of 10:

so log(1+e) becomes 10^log(1+e), which of course is just 1+e but then you need to do the same to the other side which is where I start to be not so sure:

10^log(Robs/Rtrue)/m = 10^log(1+e)
e = 10^log(Robs/Rtrue)/m - 1
e = ((Robs/Rtrue)^m-1) - 1

I am really not sure that's right but I can't seem to figure out what to do with the other side once you raise it to the 10.

If you can help me out with this I would seriously appreciate it, thanks. And thank you for the last part it has been a long time since I did logarithms so I am glad I actually did it correctly.
• Jan 7th 2010, 08:51 PM
TKHunny
$\displaystyle \frac{1}{m}log\left(\frac{R_{obs}}{R_{true}}\right ) = log(1+e)$

$\displaystyle log\left(\left[\frac{R_{obs}}{R_{true}}\right]^{\frac{1}{m}}\right) = log(1+e)$

$\displaystyle \left[\frac{R_{obs}}{R_{true}}\right]^{\frac{1}{m}} = 1+e$

Notice, please, how this demonstration did NOT require me to disclose the base of the logarithms I used.

log(a^b) = b*log(a) Is works both ways.

log(a*b) = log(a) + log(b) It works both ways.