Question

How do I solve log4y+log2y=12 ?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

Thanks guys

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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