What are the solutions of the given equation?

Question

#4^(x-1) - 3.2^(x-1) + 2 = 0#

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Answer 1 · 2017-10-07T09:19:23

#x = 1" "# or #" "x = 2#

We have:

#0 = 4(4^(x-1)-3*2^(x-1)+2)#

#color(white)(0) = (2^x)^2-6(2^x)+8#

#color(white)(0) = (2^x-2)(2^x-4)#

So:

#2^x = 2" "# or #" "2^x = 4 = 2^2#

Hence:

#x = 1" "# or #" "x = 2#

Answer 2 · 2017-10-07T09:22:47

#x=2" or "1#

Le t #2^(x-1)=t#, then #4^(x-1)-3*2^(x-1)+2=0# becomes

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

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

i.e. either #x=2# or #x=1#

therefore either #2^(x-1)=2=2^1# i.e. #x-1=1=>x=2#

or #2^(x-1)=`=2^0# i.e. #x-1=0=>x=1#

Hence #x=2" or "1#