Math Help - Mass Bias Calculation

1. 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

2. Well, how can we tell you if you are right if you don't present anything.

Logarithms are not tricky. Stop thinking that.

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

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

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

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

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

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

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

You're not going to make me do ALL the work, are you?

3. 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.

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

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

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

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.