Solving a 'Harvard' University entrance exam question

[00:00:00]Hello Friends find the value of 'h' If h^32=2^h let's have a solution to solve this problem, take power (1/h) on both sides (h^32)^(1/h)=(2^h)^(1/h) (a^m)^n=a^mn then It will be h^(32x1/h)=2^(hx1/h) where 'h' cancels h^(32/h)=2 in the next step, take power (1/32) on both sides (h^(32/h))^(1/32)=2^(1/32) where 32 cancels h^(1/h)=2^(1/32) which is same as h^(1/h)=2^(1x1/32) h^(1/h)=2^(2/2x1/32) h^(1/h)=(2^2)^(1/2x1/32) h^(1/h)=4^(1/64) h^(1/h)=4^(2/2x1/64) h^(1/h)=(4^2)^(1/2x1/64) h^(1/h)=16^(1/128) h^(1/h)=16^(2/2x1/128) h^(1/h)=(16^2)^(1/2x1/128) h^(1/h)=(256)^(1/256) so, by comparing, we get the value of 'h=256' in the next step, I'm going to verify h^32=2^h put the value of h=256 256^32=2^256 (2^8)^32=2^256 2^256=2^256 L.H.S=R.H.S which shows that the value of 'h=256' satisfies this equation of 'h^32=2^h' thanks for watching this video please subscribe this channel to get the notification of my new videos ok bye