SOLUTION: Solve 3^2x=7(3^x)-12 and check for extraneous roots. I know this is a quadratic function. So everything to one side : 3^2x-7(3^x)+12 and applied the quadratic formula. my z

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Solve 3^2x=7(3^x)-12 and check for extraneous roots. I know this is a quadratic function. So everything to one side : 3^2x-7(3^x)+12 and applied the quadratic formula. my z      Log On


   

Question 693478: Solve 3^2x=7(3^x)-12 and check for extraneous roots.

I know this is a quadratic function. So everything to one side :
3^2x-7(3^x)+12 and applied the quadratic formula.
my zeros were x=4 or x=3, however the answers are 1, and 1.26. So I Don't know how to solve this.
Answer by mouk(232) About Me  (Show Source):
You can put this solution on YOUR website!
You were very close to solving this.
+3%5E%282x%29=7%283%5Ex%29-12+
+3%5E%282x%29-7%283%5Ex%29%2B12+=+0+
This is quadratic in +3%5Ex+ and not +x+
so you can factorize (using the roots you stated)
+%283%5Ex-3%29%283%5Ex-4%29=0+
so the solutions are +3%5Ex=4+ and +3%5Ex=3+
or, by taking logs to base 10 +xlog3+=log4+ and +xlog3=log3+
giving x=1.26185... and x=1 as stated.