solve this equation for m0
vm=u(lnm0-lnms)-gtf
Add gtf to both sides then divide vm+gtf by u.
Then add lnms so ((vm+gtf)/u)+lnms=lnmo
Then to get rid of the natural log (ln) you put both sides in the exponent with base e so e^(((vm+gtf)/u)+lnms)=e^(lnmo)
And e and ln cancel out on the right side to leave you with
mo=e^(((vm+gtf)/u)+lnms)
If you still have problems understanding how I found the answer, ask your teacher and I am sure they will help you understand :)