Respuesta :
The general equation is:
[tex]\frac{b^{2k}\cdot b^{3k}}{b^{4k}}=b^k[/tex]In the firs case, b = 5 and k = 1.
[tex]\begin{gathered} \frac{5^{2\cdot1}+5^{3\cdot1}}{5^{4\cdot1}}=5^1 \\ \frac{5^2+5^3}{5^4}=5^{} \end{gathered}[/tex]In the second case, b = 9 and k = 2.
[tex]\begin{gathered} \frac{9^{2\cdot2}+9^{3\cdot2}}{9^{4\cdot2}}=9^2 \\ \frac{9^4+9^6}{9^8}=9^2 \end{gathered}[/tex]In the third case, b = 2 and k = 3.
[tex]\begin{gathered} \frac{2^{2\cdot3}+2^{3\cdot3}}{2^{4\cdot3}}=2^3 \\ \frac{2^6+2^9}{2^{12}}=2^3 \end{gathered}[/tex]In the fourth case, b = 3 and k = 4.
[tex]\begin{gathered} \frac{3^{2\cdot4}+3^{3\cdot4}}{3^{4\cdot4}}=3^4 \\ \frac{3^8+3^{12}}{3^{16}}=3^4 \end{gathered}[/tex]