Question

Sejam: y = [(1 + logp m). logpm c] - logp c e x= logp cm - log√(m&p)C, então x- y, vale:

Answer 1

 y = [(1 + log[p] M) . log[pm] C] - log[p] C 

y = [(1 + log[p] M) .( log[p]C/log[p] PM)] - log[p] C 

y = [(1 + log[p] M) .( log[p]C/log[p] P + log[p]M)] - log[p] C 

y = [(1 + log[p] M) .( log[p]C/1 + log[p]M)] - log[p] C 

y = [(1 + log[p] M) .( log[p]C)/(1 + log[p]M)] - log[p] C 

y = log[p] C - log[p] C 

y = 0 
----------------------------- 

x = log[p] C^(m) - log[m√p] C 

x = log[p] C^(m) - (log[p] C/log[p] m√P) 

x = log[p] C^(m) - [log[p] C/log[p] P^(1/m)] 

x = log[p] C^(m) - [log[p] C/(1/m).log[p] P] 

x = log[p] C^(m) - [log[p] C/(1/m)] 

x = log[p] C^(m) - [(log[p] C).(m/1)] 

x = log[p] C^(m) - m.log[p] C 

x = log[p] C^(m) - log[p] C^(m) 

x = 0 
----------------------------------- 

x - y = 0 - 0 

x - y = 0