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

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Answer 1 · 2017-11-23T13:44:27

#x=0, 1#

#log(4^x+2)-xlog2=log3#

using

#logx^n=nlogx#

#log(4^x+2)-log2^x=log3#

using

#logX-logY=log(X/Y)#

#log((4^x+2)/2^x)=log3#

because the logs are the same base

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

#=>4^x+2=3*2^x#

#4^x-3*2^x+2=0#

#2^(2x)-3*2^x+2=0#

we have a quadratic in #2^x# and it factorises

#(2^x-1)(2^x-2)=0#

#:.2^x=1=>x=0#

or

#2^x=2=>x=1#