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An alternative approach, assuming this is doable enough by hand: $$\left(\frac{7}{9}\right)^3 = \frac{343}{729} < \frac{1}{2}$$ Then: $$\frac{7^{19}}{9^{17}} = \left(\frac{7}{9}\right)^{18} \; \frac{9}{7} \; 7^2 = \left( \left(\frac{7}{9}\right)^3 \right)^6 \; 63 < \left( \frac{1}{2} \right)^6 63 = \frac{63}{64} < 1$$ So $7^{19}<9^{17}$ after fairly easy and straightforward calculations.

An alternative approach, assuming this is doable enough by hand: $$\color{blue}{\left(\frac{7}{9}\right)^3} = \frac{343}{729} \color{blue}{< \frac{1}{2}}$$ Then: $$\color{red}{\frac{7^{19}}{9^{17}}} = \left(\frac{7}{9}\right)^{18} \; \frac{9}{7} \; 7^2 = \left( \color{blue}{\left(\frac{7}{9}\right)^3} \right)^6 \; 63 \color{blue}{<} \left(\color{blue}{ \frac{1}{2}} \right)^6 63 = \frac{63}{64} \color{red}{< 1}$$ So $\color{red}{7^{19}<9^{17}}$ after fairly easy and straightforward calculations.

An alternative approach, assuming this is doable enough by hand: $$\left(\frac{7}{9}\right)^3 = \frac{343}{729} < \frac{1}{2}$$ Then: $$\frac{7^{19}}{9^{17}} = \left(\frac{7}{9}\right)^{18} \; \frac{9}{7} \; 7^2 = \left( \left(\frac{7}{9}\right)^3 \right)^6 \; \frac{9}{7} \; 7^2 < \left( \frac{1}{2} \right)^6 63 = \frac{63}{64} < 1$$ So $7^{19}<9^{17}$ after fairly easy and straightforward calculations.

An alternative approach, assuming this is doable enough by hand: $$\left(\frac{7}{9}\right)^3 = \frac{343}{729} < \frac{1}{2}$$ Then: $$\frac{7^{19}}{9^{17}} = \left(\frac{7}{9}\right)^{18} \; \frac{9}{7} \; 7^2 = \left( \left(\frac{7}{9}\right)^3 \right)^6 \; 63 < \left( \frac{1}{2} \right)^6 63 = \frac{63}{64} < 1$$ So $7^{19}<9^{17}$ after fairly easy and straightforward calculations.

Alternative, assuming this is doable enough by hand: $$\left(\frac{7}{9}\right)^3 = \frac{343}{729} < \frac{1}{2}$$ Then: $$\frac{7^{19}}{9^{17}} = \left(\frac{7}{9}\right)^{17}7^2 = \left( \left(\frac{7}{9}\right)^3 \right)^6 \; \frac{9}{7} \; 7^2 < \left( \frac{1}{2} \right)^6 63 = \frac{63}{64} < 1$$ So $7^{19}<9^{17}$.

An alternative approach, assuming this is doable enough by hand: $$\left(\frac{7}{9}\right)^3 = \frac{343}{729} < \frac{1}{2}$$ Then: $$\frac{7^{19}}{9^{17}} = \left(\frac{7}{9}\right)^{18} \; \frac{9}{7} \; 7^2 = \left( \left(\frac{7}{9}\right)^3 \right)^6 \; \frac{9}{7} \; 7^2 < \left( \frac{1}{2} \right)^6 63 = \frac{63}{64} < 1$$ So $7^{19}<9^{17}$ after fairly easy and straightforward calculations.