How do you solve #\log (\frac{4^{x}+2}{2^{x}})=\log 3#?

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Answer 1 · 2017-11-23T21:55:11

#x =0 and 1#.

We can see:

#(4^x + 2)/(2^x) = 3#

#4^x +2 = 3(2^x)#

#(2^2)^x + 2^1 = 3(2^x)#

If we let #t = 2^x#, then:

#t^2 + 2 = 3t#

#t^2 - 3t + 2= 0#

#(t - 2)(t - 1) = 0

#t = 1 or 2#

We now must reverse our substitutions.

#2^x = 1 -> x= 0#
#2^x = 2 -> x = 1#

Hopefully this helps!