Question

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

Answer 1

[2^2x +2^x-12=0\]

Answer 2

put 2^x = y
i think you mean

\[2^{2x} +2^x-12=0\]

Answer 3

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

Answer 4

Didn't think of doing it that way.. I'll try it now

Answer 5

ask if you get stuck anywhere..

Answer 6

I get -7/2 and 3...

Answer 7

y^2+y-12=0
(y+4)(y-3)=0
y=-4,3
what you got -7/2 and 3 for ? x ?

Answer 8

Yup

Answer 9

did you get how y=-4,3 ?

Answer 10

The answer is supposed to be
ln3/ln2

Answer 11

Yes, that part i did.. and i plugged those Y's into X

Answer 12

\(2^x=3 \\ \ln 2^x=\ln 3 \\ x \ln 2 = \ln 3 \\ x= \ln3/ \ln 2\)

Answer 13

Just saw where I messed up.. I pugged the Y's into 2^x...
Thanks!!

Answer 14

welcome ^_^