Question

Sendo a=(2^48+4^22-2^46)/(2^2.4^21 ),obtenha o valor de (a/13)^(-1).

Answer 1

$a=\frac{2^{48}+4^{22}-2^{46}}{2^2.4^{21}} = \frac{2^{48}+2^{2^{22}}-2^{46}}{2^2.2^{2^{21}}} = \frac{2^{48}+2^{2.22}-2^{46}}{2^2.2^{2.21}} = \frac{2^{48}+2^{44}-2^{46}}{2^2.2^{42}} = \frac{2^{48}+2^{44}-2^{46}}{2^{2+42}} = \frac{2^{48}+2^{44}-2^{46}}{2^{44}} = \frac{2^{44}.(2^{4}+1-2^{2})}{2^{44}} = 2^{4}+1-2^{2} = 16+1-4 = 13$

Passos:

1-) 4 = 2²

2-) $a^{b^c} = a^{b.c}$

3-) resolvi os b.c do passo anterior

4-) $a^n.a^m = a^{n+m}$

5-) Resolvi n+m do passo anterior

6-) Coloquei $2^{44}$ em evidência

7-)  $2^{44}$ / $2^{44}$  = 1

8-) Resolvi o que restou

$(\frac{a}{13})^{-1} = \frac{13}{a} = \frac{13}{13} = 1$