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I need help with this log function... 2^2x +^x-12=0

Mathematics
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[2^2x +2^x-12=0\]
put 2^x = y i think you mean \[2^{2x} +2^x-12=0\]
\[(2^{x})^2 +2^x-12=0 \\ 2^x=y \implies y^2+y-12=0\] can you solve this quadratic in y ?

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Didn't think of doing it that way.. I'll try it now
ask if you get stuck anywhere..
I get -7/2 and 3...
y^2+y-12=0 (y+4)(y-3)=0 y=-4,3 what you got -7/2 and 3 for ? x ?
Yup
did you get how y=-4,3 ?
The answer is supposed to be ln3/ln2
Yes, that part i did.. and i plugged those Y's into X
\(2^x=3 \\ \ln 2^x=\ln 3 \\ x \ln 2 = \ln 3 \\ x= \ln3/ \ln 2\)
Just saw where I messed up.. I pugged the Y's into 2^x... Thanks!!
welcome ^_^

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