I'm trying to Isolate DR in the function below. Was wondering if I got it correct.

$(1 + DR)^y$ = $(1 + N/C)^C$

My answer

$$Dr = e^{\ln(1 + N/C)^C \over y}$$

Sorry about that last line. eveyrthing in the bracket after exponential is making the exponential raised by that.

  • 1
    $\begingroup$ Raise both sides to the power $\frac{1}{y}$ and you are almost finished. No log stuff needed. $\endgroup$ Jan 8, 2015 at 19:35
  • $\begingroup$ Is this a typo? "Nint" instead of "N"? $\endgroup$
    – MPW
    Jan 8, 2015 at 19:40

1 Answer 1


If $$(1+DR)^y = \left(1+\tfrac NC\right)^C$$ then $$DR = \left(1+\tfrac NC\right)^{\frac Cy} - 1$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .