How do I solve log4y+log2y=12 ?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

How do I solve log4y+log2y=12 ?

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question