Asked by presh
solve this equation for m0
vm=u(lnm0-lnms)-gtf
vm=u(lnm0-lnms)-gtf
Answers
Answered by
#1Student
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 :)
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 :)
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