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$$\frac{\ln(T+1)-\ln(V-S+1)}{\ln(V-S+1)}=\frac{1}{K}-1$$

What are the steps if I want to solve for $V$?

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3 Answers 3

up vote 4 down vote accepted

Hint: Get rid of the second term of the numerator on the LHS as -1 and move to RHS. A little algebra will leave $\ln(V-S+1)$ on the RHS and all else on the LHS. Then exponentiate both sides.

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Set $x = \ln(V-S+1)$ and solve the equation (which is linear in $x$).

Then use $V = e^x + S - 1$.

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+1: This is so much better than the answer I was about to post! –  Eric Naslund Feb 24 '11 at 19:13

Note that the equation involves the composition of three invertible functions $\rm\ f(g(h(x))) = 0\ $ where $\rm\:f\:$ has fractional linear form $\rm\ f(x) = (a\ x + b)/(c\ x + d)\:,\: $ and $\rm\ g(x) = \ln\: x\:,\:$ and $\rm\ h(x) = x - S - 1\:.\:$ Therefore, upon applying inverses we conclude that$\rm\ x = h^{-1}(g^{-1}(f^{-1}(0)))\:.\:$ That's effectively what the other solutions do. The same method solves (inverts) any composition of invertible functions.

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