Log Prove Without Calculator

[nycmathguy]

 

[MathLover1]

1/log(2,pi)+log(5,pi)>2

use the rule:
log(b,x) = log(k,x)/log(k,b)

log(2,pi)= log(pi,pi)/log(pi,2)............log(pi,pi)=1
log(2,pi)= 1/log(pi,2).
log(5,pi)= 1/log(pi,5)


1/( 1/log(pi,2))+(1/log(pi,5))>2

log(pi,2)+log(pi,5))>2

log(pi,2*5)>2

log(pi,10)>2->true

 

[nycmathguy]

1/log(2,pi)+log(5,pi)>2

use the rule:
log(b,x) = log(k,x)/log(k,b)

log(2,pi)= log(pi,pi)/log(pi,2)............log(pi,pi)=1
log(2,pi)= 1/log(pi,2).
log(5,pi)= 1/log(pi,5)


1/( 1/log(pi,2))+(1/log(pi,5))>2

log(pi,2)+log(pi,5))>2

log(pi,2*5)>2

log(pi,10)>2->true

Nice job like always. It is easier, however, to read your answers on paper just like I post my threads.