Basic Variable Isolation

Question

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.

Raise both sides to the power $\frac{1}{y}$ and you are almost finished. No log stuff needed. — André Nicolas, Jan 08 '15 at 19:35

score 1 · Accepted Answer · answered Jan 08 '15 at 19:37

1

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

answered Jan 08 '15 at 19:37

MPW

1 Answers1