Question

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

Answer 1

Answer:

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