# V=B-q/k In(r-1): Make r the subject.

• Oct 6th 2010, 10:45 AM
manfred
V=B-q/k In(r-1): Make r the subject.
hi

i have a equation which i dont know how to solve it, as i have never seen this equation before !!

question

given : V=B-q/k In(r-1)

make r subject of the formula ?

could you please explain what they want exactly

thanks
• Oct 6th 2010, 12:26 PM
mr fantastic
Quote:

Originally Posted by manfred
hi

i have a equation which i dont know how to solve it, as i have never seen this equation before !!

question

given : V=B-q/k In(r-1)

make r subject of the formula ?

could you please explain what they want exactly

thanks

They want you to solve for r in terms of the other pronumerals. I suggest that you first attempt to make ln(r-1) the subject.

By the way, your expression is ambiguous. Is it meant to be V=B-q/[k In(r-1)] or V=[B-q/k] In(r-1) ....?? Or V=(B-q)/[k In(r-1)] or V=B-(q/[k In(r-1)]) etc. etc. Please use bracketes, or take the couple of minutes to learn how to use latex (click on the appropriate link in my signature).
• Oct 6th 2010, 12:34 PM
manfred
Quote:

Originally Posted by mr fantastic
They want you to solve for r in terms of the other pronumerals. I suggest that you first attempt to make ln(r-1) the subject.

By the way, your expression is ambiguous. Is it meant to be V=B-q/[k In(r-1)] or V=[B-q/k] In(r-1) ....?? Or V=(B-q)/[k In(r-1)] or V=B-(q/[k In(r-1)]) etc. etc. Please use bracketes, or take the couple of minutes to learn how to use latex (click on the appropriate link in my signature).

yes i meant V=[B-q/k] In(r-1)

could you please show me the solutions to it then i can understand how and where its coming from in solving it please thanks
• Oct 6th 2010, 03:04 PM
skeeter
Quote:

Originally Posted by manfred
yes i meant V=[B-q/k] In(r-1)

could you please show me the solutions to it then i can understand how and where its coming from in solving it please thanks

$\displaystyle \displaystyle V=\left(B - \frac{q}{k}\right) \cdot \ln(r-1)$

$\displaystyle \displaystyle \frac{V}{B - \frac{q}{k}} = \ln(r-1)$

$\displaystyle \displaystyle e^{\frac{V}{B - \frac{q}{k}}} = r - 1$

$\displaystyle \displaystyle 1 + e^{\frac{V}{B - \frac{q}{k}}} = r$
• Oct 7th 2010, 03:55 AM
HallsofIvy
Quote:

Originally Posted by manfred
hi

i have a equation which i dont know how to solve it, as i have never seen this equation before !!

I hope you do NOT mean that the only equations you can solve are those that you have seen before!

Quote:

question

given : V=B-q/k In(r-1)

make r subject of the formula ?

could you please explain what they want exactly

thanks
By the way, it is "ln" (small L) NOT "In", for natural logarithm.
• Oct 7th 2010, 04:21 AM
Wilmer
Quote:

Originally Posted by manfred
yes i meant V=[B-q/k] In(r-1)

Well, make your life easier by making x = B - q/k; then:
Ln(r - 1) = V / x
r - 1 = e^(V / x)