anonymous
  • anonymous
Which logarithmic graph can be used to approximate the value of y in the equation 6y = 12? I WILL MEDAL/FAN! PLEASE HELP!!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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.
chestercat
  • chestercat
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anonymous
  • anonymous
1. http://assets.openstudy.com/updates/attachments/53a8acb5e4b0ffdda15e7281-xiphoidx-1403563342568-1.png
anonymous
  • anonymous
2. http://assets.openstudy.com/updates/attachments/53a8acb5e4b0ffdda15e7281-xiphoidx-1403563342359-2.png 3. http://assets.openstudy.com/updates/attachments/53a8acb5e4b0ffdda15e7281-xiphoidx-1403563342187-3.png 4. http://assets.openstudy.com/updates/attachments/53a8acb5e4b0ffdda15e7281-xiphoidx-1403563341991-4.png
anonymous
  • anonymous
@mathmate

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anonymous
  • anonymous
Do you understand it?
anonymous
  • anonymous
Should we call for backup?
mathmate
  • mathmate
First you would solve the equation \(6^y=x\) take log on both sides \(log 6^y = log x\) use the properties of exponents to change it to \(y~log 6 = log x\) or \(\large y=\frac{log(x)}{log6}\) Since the given equation has x=12, substitute x=12 in the above equation and solve for y. Use the point (12,y) to select the graph required.
anonymous
  • anonymous
Ok, hold on:)
anonymous
  • anonymous
So, it's either A or B yes?
Loser66
  • Loser66
yes
anonymous
  • anonymous
I got A :)
Loser66
  • Loser66
yup
anonymous
  • anonymous
Thanks :)

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