## toadytica305 I need help with this log function... 2^2x +^x-12=0 one year ago one year ago

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

2. hartnn

put 2^x = y i think you mean $2^{2x} +2^x-12=0$

3. hartnn

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

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

5. hartnn

ask if you get stuck anywhere..

I get -7/2 and 3...

7. hartnn

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

Yup

9. hartnn

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

12. hartnn

$$2^x=3 \\ \ln 2^x=\ln 3 \\ x \ln 2 = \ln 3 \\ x= \ln3/ \ln 2$$