Question 48

If $$R_c=m\times\ln\left(1+\ \frac{\ R_m}{m}\right)$$ then $$R_m$$ is equal to

$$\ \frac{\ R_c}{m}=\ln\left(1+\frac{R_m}{m}\right)$$

$$\ \frac{\ R_m}{m}+\ 1\ =\ e^{\ \frac{\ R_c}{m}}$$

$$\ \frac{\ R_m}{m}\ =\ e^{\ \frac{\ R_c}{m}}-1$$

$$R_m = m\left(e^{\frac{R_c}{m}} - 1 \right)$$

C is the correct answer.

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