Question

if x=loga base 2a, y= 2a base 3a & z= log 3a base 4a,show that , xyz+1= 2yz

Answer 1

x=log (a base 2a)=loga/(log2+loga)
y=log (2a base 3a)=(log2+loga)/(log3+loga)
z=log (3a base 4a)=(log3+loga)/(log4+loga)

now
LHS=xyz+1
put above x, y, z value
={(loga)/(log2+loga)}{(log2+loga)/(log3+loga)}{(log3+loga)/(log4+loga)}+1

=loga/(log4+loga)+1

=loga/(log4a)+1

=log (a base 4a)+1

RHS=2yz

=2 {(log2+loga)/(log3+loga)}{(log3+loga)/(log4+loga)}

=2 (log2+loga)/(log4+loga)

=2log2a/log4a

=log4a^2/log4a

=log4a.a/log4a

=log4a/log4a+loga/log4a

=1+loga/log4a

=log (a base 4a ) +1

hence LHS=RHS

Answer 2

Answer:

Hey mate your answer is here