natural logs

[jpd5184]

1. The problem statement, all variables and given/known data

(R-V)/ln S = U/ e-k

solve for R

these are just rudimentary letters, they dont mean anything

3. The attempt at a solution

i get R-V= (u / e-k)ln s
then add V to both sides and get:

R= [(u/e-k)ln s] + v

i then have to get rid of the natural log so do e^(something)

[Outlined]

why do you want to get rid of it ?

[jpd5184]

i don't know just making a suggestion. why wouldn't you get rid of the natural log?

[Mark44]

You want to solve for R. Step 1 is to multiply by sides by ln(S).

The answer will still have the ln term in it.

[Mentallic]

If you wanted to remove that log, then you would take the exponential of both sides to get

e^R=e^{\frac{u}{e-k}ln(s)+v}

e^R=\left(e^{ln(s)}\right)^{\frac{u}{e-k}}e^v

e^R=s^{\frac{u}{e-k}}e^v


Ok so we got rid of the log, but we haven't done what the original question asked of us, to solve for R. So as others have said, R will be in terms of log(s).

[HallsofIvy]

1. The problem statement, all variables and given/known data

(R-V)/ln S = U/ e-k

solve for R

these are just rudimentary letters, they dont mean anything

3. The attempt at a solution

i get R-V= (u / e-k)ln s
then add V to both sides and get:

R= [(u/e-k)ln s] + v
You said you wanted to solve for R. This is solved for R.

i then have to get rid of the natural log so do e^(something)
But then it would not be solved for R.