Question

Sejam y=[(1 + log de p elevado m) . log de p . m elevado c] - log p elevado c e x= log p elevado c elevado m - log de indice m raiz dep elevado c.Então, x - y vale

Answer 1

para y:

$y= [ (1+ log p^{m}) . log pm^{c}] - log p^{c}$

$= [ (log p^p+ log p^m). \frac{logp^c}{logp^pm}] - logp^c =$

= $logp^pm + \frac{logp^c}{logp^pm} - logp^c$

=$logp^c-logp^c = 0$

Para o x:

$x= log p^{c^{m}} - log \sqrt[m]{p}^c$

=$m. logp^c - log p ^ \frac{1}{m} ^{c}$

= $m.log p^c - m.logp^c$

=0

Portanto, x-y = 0-0 = 0.$\bigstar \bigstar \bigstar \bigstar \bigstar \bigstar \bigstar \bigstar \bigstar \bigstar \bigstar$