If #64^x-1=16^(x/2)#, find #x#?

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Answer 1 · 2017-11-09T06:55:20

#x=3/2#

It is assumed here that we are given #64^(x-1)=16^(x/2)# and not #64^x-1=16^(x/2)#

#64^(x-1)=16^(x/2)#

or #ln64^(x-1)=ln16^(x/2)#

or #(x-1)ln64=(x/2)ln16#

or #(x-1)(2/x)=ln16/ln64#

or #(x-1)1/x=ln16/(2ln64)#

or #1-1/x=ln2^4/(2ln2^6)#

or #-1/x=(4ln2)/(12ln2)-1#

or #-1/x=1/3-1=-2/3#

or #1/x=2/3#

i.e. #x=3/2#