Question

if x=(log base 2a).a, y=(log base 3a).2a , z=(log base 4a).3a ; prove that xyz+1=2yz

Answer 1

We have:log2a(a)=xlog3a(2a)=ylog4a(3a)=zxyz=log2a(a) log3a(2a) log4a(3a)By base change formula we have:logyx=logzxlogzy, we get:xyz=logc(a)logc(2a).logc(2a)logc(3a).logc(3a)logc(4a)=logc(a)logc(4a)⇒xyz=log4a(a)....(i)2yz=2log3a(2a) log4a(3a)=2logc(2a)logc(3a).logc(3a)logc(4a)=2 logc(2a)logc(4a)⇒2yz=2 log4a(2a)= log4a(2a)2....(As log(mn)=nlog(m))⇒2yz= log4a(4a2)....(ii)(i)−(ii), we getxyz−2yz=log4a(a)− log4a(4a2)=−{ log4a(4a2)−log4a(a)}=−log4a(4a2a)  [log(a)−log(b)=log(ab)]=−log4a(4a)=−1Hencexyz−2yz=−1...
xyz+1=2yz
hence proved