Chapter 2 - Nonlinear Functions - 2.5 Logarithmic Functions - 2.5 Exercises - Page 101: 89

$\frac{s}{n}=2^{ \frac{C}{B}}-1$

$C=B\log_2(\frac{s}{n}+1)$ $B\log_2(\frac{s}{n}+1)=C$ $\log_2(\frac{s}{n}+1)=\frac{C}{B}$ In exponential form $2^{ \frac{C}{B}}=\frac{s}{n}+1$ $\frac{s}{n}+1=2^{ \frac{C}{B}}$ $\frac{s}{n}=2^{ \frac{C}{B}}-1$