anonymous 5 years ago How do I solve log4y+log2y=12 ?

1. anonymous

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

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. anonymous

Thanks guys

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

$y=\frac{\sqrt3}{\sqrt2}=\frac{\sqrt6}{2}$ -- 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").

10. anonymous

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