Solving a 'Stanford' University entrance exam | t=?

Ajouté : 2026-05-11

[00:00:00]Hello Friends find the value of 't' If 3^t.3^t=36 let's have a solution so, we have a problem of 3^t.3^t=36 as x.x=x^2 then It will be (3^t)^2=36 (a^m)^n=a^mn 3^(2t)=36 we know 36=6x6=6^2 then we can write it as 3^(2t)=6^2 now, take 'log' on both sides log3^(2t)=log6^2 we know that logm^n=nlogm we have 2tlog3=2log6 divide by '2log3' on both sides 2tlog3/2log3=2log6/2log3 2 and log3 cancels we get t=log6/log3 since 6=3x2 then t=log(3x2)/log3 next, logmn=logm+logn t=(log3+log2)/log3 separate it into fractions t=log3/log3+log2/log3 log3 cancels as we know loga/logb=logb(a) t=1+log3(2) which is our the value of 't' in the next step, I'm going to verify 3^t.3^t=36 as by solving 3^t.3^t=36, we get 3^(2t)=36 so this is same as 3^(2t)=36 now, put the value of 't' 3^2(1+log3(2))=36 3^(2+2log3(2))=36 since nlogm=logm^n then It will be 3^(2+log3(2^2))=36 3^(2+log3(4))=36 since a^(m+n)=a^m.a^n 3^2.3^log3(4)=36 as we know a^loga(b)=b 9.4=36 36=36 you can see L.H.S=R.H.S which shows that the value of t=1+log3(2) satisfies this equation of 3^t.3^t=36 thanks for watching this video please subscribe this channel to get the notification of my new videos ok bye