Solving a 'Harvard' University entrance exam question

[00:00:00]Hello Friends find the value of 't' If 8^t/64=6 let's have a solution in this problem. '8' is same as '2^3', 64' is same as '2^6' and '6' is same as 2x3 as we know (a^m)^n=a^(mn) & a^m/a^n=a^(m-n) then this equation can be written as 2^(3t-6)=2x3 divide by '2' on both sides, 2^(3t-6)/2=2x3/2 where '2' cancels 2^(3t-6-1)=3 2^(3t-7)=3 take 'log' on both sides log(2^(3t-7))=log3 since logm^n=nlogm, apply this formula (3t-7)log2=log3 in the next step, divide by 'log2' on both sides, 'log2' cancels, 3t-7=log3/log2 as loga/logb=logb(a), apply formula we have 3t-7=log2(3) 3t=7+log2(3) divide by '3' on both sides, 3t/3=(7+log2(3))/3 where '3' cancels we get the value of 't=(7+log2(3))/3' in the next step, I'm going to verify, 8^t/64=6 8^(7+log2(3)/3)/64=6 '8' is same as '2^3', 2^3(7+log2(3)/3)/64=6 here, '3' cancels, we get 2^(7+log2(3))/64=6 since a^(m+n)=a^m.a^n using this, we have 2^7.2^log2(3)/64=6 a^(loga(b))=b, using this formula 2^7=128, 2^(log2(3))=3 128.3/64=6 64x2=128 2.3=6 6=6, you can see L.H.S=R.H.S which shows that the value of 't=(7+log2(3))/3' satisfies this equation of '8^t/64=6' thanks for watching this video please subscribe this channel to get the notification of my new videos ok bye