Determining exponents in a rate law with different data
If you've correctly determined exponent $m$, then to ascertain $n$ you can proceed by first dividing rate equation $(3)$ by rate equation $(2)$ (note that the rate constant $k$ cancels immediately and we have that $m = 1$):

$$
\dfrac{\text{rate}_3}{\text{rate}_2} = \dfrac{\ce{[C2H4Br2]}_3}{\ce{[C2H4Br2]}_2} \bigg(\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}\bigg)^n
$$

Rearranging this equation, one obtains:

$$
\dfrac{\text{rate}_3}{\text{rate}_2} \cdot \dfrac{\ce{[C2H4Br2]}_2}{\ce{[C2H4Br2]}_3} = \bigg(\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}\bigg)^n
$$

Now, to isolate $n$, take the logarithm (in any base of your choice, though the book arbitrarily uses the natural log, and I'll do likewise out of concern for consistency) of both sides of the equation, and remember that one of the basic properties of logarithms is that $\log_b x^n = n\log_b x$:

$$
\ln\dfrac{\text{rate}_3 \ce{[C2H4Br2]}_2}{\text{rate}_2 \ce{[C2H4Br2]}_3} = \ln\bigg(\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}\bigg)^n = n\ln\dfrac{\ce{[I^{-}]}_3}{\ce{[I^{-}]}_2}
$$

Solving for $n$, one finds:

$$
\frac{\ln\dfrac{\text{rate}_3 \ce{[C2H4Br2]}_2}{\text{rate}_2 \ce{[C2H4Br2]}_3}}{\ln\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}} = n
$$

Which is the final equation from your textbook's solutions manual.