Can you help me solve this

Question

log6x=               1                

                      1     +                  1       

                log2x              log 3 (x)

Answer 1

Let y=log_a(b), so b=a^y; let z=log_b(a), so a=b^z=(a^y)^z=a^yz; therefore, equating exponents, yz=1, y=1/z, hence, log_a(b)=1/log_b(a), so log_x(6)=1/log₂(x)+1/log₃(x)=log_x(2)+log_x(3)=log_x(6). So this is an identity, true for all x>1.