log_bM^r=rlog_bM
guys how to prove this ? if you have any idea please help me. thanks.

@ comingsoon,

umm! thanks Black Panther .. ...... appreciated .

Not sure if this is the way to solve , however trying is not bad.

 

 

log_bM^r=rlog_bM

log_bM^r=log_b  (M1+M2+M3...+Mr)   //m1,m2,m3 =M ; 1,2,3 represents index  

        =log_bM1+log_bM2+log_bM3+....log_bMr

M1, M2, M3 ..Mr represents same variable M there fore the next step is just arithmetic addition, i.e addition of r variable =rx

     =rlog_bM

 

 

 

@Black_Panther, Did you only copy paste the solution?
How would it be:
log _a xⁿ = log _a a^mn
log _a xⁿ = mn log _a a
Without actually prooving log_a xⁿ = nlog_a x ? It is using the same thing which was actually asked to prove.

 

log_bM^r=log_b  (M1+M2+M3...+Mr)   //m1,m2,m3 =M ; 1,2,3 represents index  

was supposed to be

 

log_bM^r=log_b  (M1.M2.M3....Mr)   //m1,m2,m3 =M ; 1,2,3 represents index