anonymous
  • anonymous
Simplify log(base a)1-log(base a)a^b
Mathematics
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.

anonymous
  • anonymous
Simplify log(base a)1-log(base a)a^b
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
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

anonymous
  • anonymous
log(1/(a^b)) => log(a^(-b)) => -blog(a) => -b sing base is a we have -b
anonymous
  • anonymous
Remember that by definition, for any number x, $$\log_x 1= 0,$$ since any number to the zeroth power is 1. Also, remember that, by definition for any number x, $$\log_x x^b = b.$$ That's what a logarithm is. So combining those two: $$\log_a 1 - \log_a a^b = 1 - b$$ That's right, the a doesn't matter at all!
anonymous
  • anonymous
Sorry, I that last equation was supposed to read: $$\log_a 1 - \log_a a^b = -b$$

Looking for something else?

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

More answers

anonymous
  • anonymous
Rule #1 of posting corrections: your correction will itself contain a mistake.
anonymous
  • anonymous
thanks, i couldnt put the equation thing. so the answer is just -b??? that simple?
amistre64
  • amistre64
yes, -b
anonymous
  • anonymous
but you said logx1=0 so shoudltn it be 0-b?
amistre64
  • amistre64
0 is not a "simplification"; it is an overcomplication :)
amistre64
  • amistre64
0 -b = -b ; see :)
anonymous
  • anonymous
ok :) thanks

Looking for something else?

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