Simple fraction problem from book

Question

How does the step from $$=\biggr(\frac{7}{4}\biggr)^{k-2}\biggr(\frac{11}{4}\biggl)$$ To $$=\biggr(\frac{7}{4}\biggr)^{k-2}\biggr(\frac{7}{4}\biggl)^2$$ work given $(\frac{7}{4})^2 \neq (\frac{11}{4})$

Answer 1

The expression $\displaystyle {(\frac 74)}^{k-2}(\frac {11}4) <{(\frac 74)}^{k-2}(\frac {7}4)^2$ means the left hand term is less than the right hand term. It is not equality. It is an inequality. The expression $a < b$ means that $a$ is less than $b$. The expression $a > b$ means that $a$ is greater than $b$.

And the inequality holds because $\displaystyle (\frac {11}4) < (\frac {7}4)^2$. Do you see now?

Can you see how to prove that $\displaystyle (\frac {11}4) < (\frac {7}4)^2$, or do you need help with this part too?

Answer 2

$$\left(\frac{7}{4}\right)^{k-2}\cdot \frac{11}{4}=\left(\frac{7}{4}\right)^k\left(\frac{4}{7}\right)^2\cdot \frac{11}{4}=\left(\frac{7}{4}\right)^k\frac{44}{49}$$