Solving a 'Harvard' University entrance exam question

Ajouté : 2026-05-19

[00:00:00]Hello Friends find the value of 'x' If log2(x)+log3(x)=5 let's have a solution as logb(a)=loga/logb so It will be logx/log2+logx/log3=5 'logx' can be common logx(1/log2+1/log3)=5 take LCM logx(log3+log2/log2.log3)=5 as loga+logb=logab we have logx(log(3.2)/log2.log3)=5 logx(log6/log2.log3)=5 It can be written as logx=5.log2.log3/log6 which is same as logx/log2=5.log3/log6 as loga/logb=logb(a) It will be log2(x)=5.log6(3) in the next step as we know that If loga(x)=b then x=a^b so x=2^5.log6(3) I hope you understood as x^ab=(x^a)^b we have x=(2^5)^log6(3) 2^5= 2.2=4 4.2=8 8.2=16 16.2=32 2^5=2.2.2.2.2=32 finally we get the value of x x=32^log6(3) thanks for watching this video please subscribe this channel to get the notification of my new videos and don't forget to share these videos with your classmates and friends so that they also have a benefit of it you can also visit the Playlists of this channel to learn more and more ok bye