How do you solve #2^x= 4^(x+1)#?

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Answer 1 · 2015-08-17T12:52:41

#color(blue)(x=-2#

#2^x=4^(x+1)#
We know that #4=2^2#

So the expression can be written as:
#2^x=2^(2*(x+1))#

#2^x=2^(2x+2)#

Now as the bases are equal we equate the exponents to find #x#

#x=2x+2#

#-2=2x-x#

#color(blue)(x=-2#