A community for students.
Here's the question you clicked on:
 0 viewing
toadytica305
 2 years ago
I need help with this log function...
2^2x +^x12=0
toadytica305
 2 years ago
I need help with this log function... 2^2x +^x12=0

This Question is Closed

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2put 2^x = y i think you mean \[2^{2x} +2^x12=0\]

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2\[(2^{x})^2 +2^x12=0 \\ 2^x=y \implies y^2+y12=0\] can you solve this quadratic in y ?

toadytica305
 2 years ago
Best ResponseYou've already chosen the best response.0Didn't think of doing it that way.. I'll try it now

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2ask if you get stuck anywhere..

toadytica305
 2 years ago
Best ResponseYou've already chosen the best response.0I get 7/2 and 3...

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2y^2+y12=0 (y+4)(y3)=0 y=4,3 what you got 7/2 and 3 for ? x ?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2did you get how y=4,3 ?

toadytica305
 2 years ago
Best ResponseYou've already chosen the best response.0The answer is supposed to be ln3/ln2

toadytica305
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, that part i did.. and i plugged those Y's into X

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2\(2^x=3 \\ \ln 2^x=\ln 3 \\ x \ln 2 = \ln 3 \\ x= \ln3/ \ln 2\)

toadytica305
 2 years ago
Best ResponseYou've already chosen the best response.0Just saw where I messed up.. I pugged the Y's into 2^x... Thanks!!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.