How do you solve #.5(6^x) = 2^x#?

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Answer 1 · 2016-01-13T19:35:53

#x=0.631# (3sf)

Take logs of both sides.

#log(0.5*(6^x))=log(2^x)#

#log0.5 + xlog6 = xlog2#

#log0.5=xlog2-xlog6#

#log0.5=x(log2-log6)#

#x=(log0.5)/(log2-log6)#

#x=log2/(log6-log2)#

#x=log2/log3#

#x=0.631# (3sf)