if b < -60.33268f0

1. Started with
   \[\frac{x \cdot e^{\left(y \cdot \log z + \left(t - 1.0\right) \cdot \log a\right) - b}}{y}\]
   24.2
2. Applied simplify to get
   \[\color{red}{\frac{x \cdot e^{\left(y \cdot \log z + \left(t - 1.0\right) \cdot \log a\right) - b}}{y}} \leadsto \color{blue}{\frac{\frac{x}{e^{b}}}{\frac{\frac{y}{{z}^{y}}}{{a}^{\left(t - 1.0\right)}}}}\]
   30.6
3. Applied taylor to get
   \[\frac{\frac{x}{e^{b}}}{\frac{\frac{y}{{z}^{y}}}{{a}^{\left(t - 1.0\right)}}} \leadsto \frac{\frac{x}{e^{\frac{1}{b}}}}{\frac{\frac{y}{{z}^{y}}}{{a}^{\left(t - 1.0\right)}}}\]
   2.0
4. Taylor expanded around inf to get
   \[\frac{\color{red}{\frac{x}{e^{\frac{1}{b}}}}}{\frac{\frac{y}{{z}^{y}}}{{a}^{\left(t - 1.0\right)}}} \leadsto \frac{\color{blue}{\frac{x}{e^{\frac{1}{b}}}}}{\frac{\frac{y}{{z}^{y}}}{{a}^{\left(t - 1.0\right)}}}\]
   2.0

if -60.33268f0 < b

1. Started with
   \[\frac{x \cdot e^{\left(y \cdot \log z + \left(t - 1.0\right) \cdot \log a\right) - b}}{y}\]
   12.9
2. Applied simplify to get
   \[\color{red}{\frac{x \cdot e^{\left(y \cdot \log z + \left(t - 1.0\right) \cdot \log a\right) - b}}{y}} \leadsto \color{blue}{\frac{\frac{x}{e^{b}}}{\frac{\frac{y}{{z}^{y}}}{{a}^{\left(t - 1.0\right)}}}}\]
   9.5
3. Using strategy rm
   9.5
4. Applied sub-neg to get
   \[\frac{\frac{x}{e^{b}}}{\frac{\frac{y}{{z}^{y}}}{{a}^{\color{red}{\left(t - 1.0\right)}}}} \leadsto \frac{\frac{x}{e^{b}}}{\frac{\frac{y}{{z}^{y}}}{{a}^{\color{blue}{\left(t + \left(-1.0\right)\right)}}}}\]
   9.5
5. Applied unpow-prod-up to get
   \[\frac{\frac{x}{e^{b}}}{\frac{\frac{y}{{z}^{y}}}{\color{red}{{a}^{\left(t + \left(-1.0\right)\right)}}}} \leadsto \frac{\frac{x}{e^{b}}}{\frac{\frac{y}{{z}^{y}}}{\color{blue}{{a}^{t} \cdot {a}^{\left(-1.0\right)}}}}\]
   9.4
6. Applied taylor to get
   \[\frac{\frac{x}{e^{b}}}{\frac{\frac{y}{{z}^{y}}}{{a}^{t} \cdot {a}^{\left(-1.0\right)}}} \leadsto \frac{\frac{x}{e^{b}}}{\frac{\left(y + \frac{1}{2} \cdot \left({y}^{3} \cdot {\left(\log z\right)}^2\right)\right) - {y}^2 \cdot \log z}{{a}^{t} \cdot {a}^{\left(-1.0\right)}}}\]
   4.9
7. Taylor expanded around 0 to get
   \[\frac{\frac{x}{e^{b}}}{\frac{\color{red}{\left(y + \frac{1}{2} \cdot \left({y}^{3} \cdot {\left(\log z\right)}^2\right)\right) - {y}^2 \cdot \log z}}{{a}^{t} \cdot {a}^{\left(-1.0\right)}}} \leadsto \frac{\frac{x}{e^{b}}}{\frac{\color{blue}{\left(y + \frac{1}{2} \cdot \left({y}^{3} \cdot {\left(\log z\right)}^2\right)\right) - {y}^2 \cdot \log z}}{{a}^{t} \cdot {a}^{\left(-1.0\right)}}}\]
   4.9