anonymous
  • anonymous
Help with Logarithms?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
nathan917
  • nathan917
Is there anymore to the question?
anonymous
  • anonymous
These are some problems I did before it.....

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
1. log7 + log (n - 2) = log 6n log [7(n - 2)] = log (6n) 7(n - 2) = 6n n = 14 2. log5 (16) - log5 (2t) = log5 (2) log5 (16 / 2t) = log5 (2) 16 / 2t = 2 t = 4 3. log5 m = (1/3) log5 (125) log5 (m) = log5 [(125^(1/3)] m = 125^(1/3) m = 5 4. log y = (1/4) log (16) + (1/2) log (49) log y = log [16^(1/4)] + log [49^(1/2)] =log y = log (2) + log (7) log y = log (2 . 7) y = 14 5. log6 (b^2 + 2) + log6 (2) = 2 log6 [(b^2 + 2) . 2] = 2 (b^2 + 2) . 2 = 6² b^2 + 2 = 18 b^2 = 16 b = 4 or b = - 4 6. log3 (5x + 5) - log3 (x^2 - 1) = 0 log3 [(5x + 5) / (x^2 - 1)] = 0 (5x + 5) / (x^2 - 1) = 3^0 5x + 5 = x² - 1 x² - 5x - 6 = 0 x = 6 or x = - 1 x = 6 7. log2 (x - 2) + 5 = 8 - log2 (4) log2 (x - 2) + 5 = 8 - 2 log2 (x - 2) = 1 x - 2 = 2^1 x = 4
anonymous
  • anonymous
my eyes !!!!
anonymous
  • anonymous
|dw:1365379791371:dw|
anonymous
  • anonymous
|dw:1365379863911:dw|
anonymous
  • anonymous
they are inverse of eachother

Looking for something else?

Not the answer you are looking for? Search for more explanations.