How do I solve log4y+log2y=12 ?

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

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

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

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

Thanks guys

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

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

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

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

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

Join our real-time social learning platform and learn together with your friends!