= Solution
{c}
Let $x=I_0/I_s$ and $G=\alpha_0L$. Applying the integrated amplifier law to each section gives
$$
\ln70+69x=G,\qquad
\ln\frac{150}{70}+80x=G.
$$
Eliminating $G$ gives
$$
\boxed{\frac{I_0}{I_s}=\frac1{11}\ln\frac{70^2}{150}\simeq0.31694.}
$$
As a consistency check, both amplifiers have $G\simeq26.1175$. The second section receives a larger <intensity> and adds $80I_0$, compared with $69I_0$ in the first, while its multiplicative <optical gain> is much smaller. This is the expected approach to constant extracted power in <saturation>.
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