How do I solve log4y+log2y=12 ?

Question

anonymous · Answer

log 4y + log 2y = 12
=> log ( 4y * 2y ) = 12   since log(A * B) = log A + log B
=> log 8y^2 = 12
=> 8y^2 = (10)^12       
=> y^2 = (10)^12 / 8     
=> y = 10^6 / 8^1/2
therefore,  y = 1000000 / 8^1/2

anonymous · Answer

hmm but my teacher's answer is 2
Wrong or Correct?

anonymous · Answer

I have to go with aclandt's answer.  It is my understanding that in the 3rd line of this answer, you have to use e on both sides to get the 8y^2 down, which gives you 8y^2 = e^12 which is the same as aclandt's

anonymous · Answer

get the teacher to check the answer, or  check the question again to be sure you transcribed it properly

anonymous · Answer

Thanks guys

anonymous · Answer

Could the teacher's equation be:
$$\log(4y)+\log(2y)=\log(12)$$ ?????

anonymous · Answer

Then maybe...
$$\log(8y^2)=\log(12)\Rightarrow {\log}{8}+\log{y^2}=\log(12)$$

anonymous · Answer

Whoops, I thought I could get your teacher's answer with the above equation.
I'll finish the solution anyway.
$$\log(y^2)=\log(\frac{12}{8}) $$
$$10^{\log(y^2)}=10^{({\log12/8})} \Rightarrow y^2=\frac{12}{8}$$

anonymous · Answer

$$y=\frac{\sqrt3}{\sqrt2}=\frac{\sqrt6}{2}$$
-- So I didn't get 2 either.
BTW, ulthi is right in using the base 10 to cancel the base 10 log.  Raising with [\e\] is only to invert ("undo") the natural logs (written "ln").

anonymous · Answer

ya, I was assuming natural log, i.e. base e.