Answer :

dannielle
2^x=a
4^x=(2^2)^x=(2^x)^2=a^2

a^2-10a+16=0
Δ=10^2-4*1*16=100-64=36
√Δ=6
a1=(10+6)/2=16/2=8
a2=(10-6)/2=4/2=2
2^x=8=2^3⇒ x=3 or
2^x=2=2^1⇒ x=1


There are 2 solutions : x=1 or x=3
[tex]4^x-10\cdot2^x+16=0\Rightarrow(2^2)^x-10\cdot2^x+16=0\Rightarrow(2^x)^2-10\cdot2^x+16=0[/tex]

[tex]subtitute\ 2^x=t > 0\\\\t^2-10t+16=0\\t^2-2t-8t+16=0\\t(t-2)-8(t-2)=0\\(t-2)(t-8)=0\iff t-2=0\ or\ t-8=0\\\\t=2\ or\ t=8\\\\t=2^1\ or\ t=2^3\\\\therefore\ 2^x=2^1\ or\ 2^x=2^3\\\\Answer:\boxed{x=1\ or\ x=3}[/tex]

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