UMULAS
  • UMULAS
What is the solution to the equation log 2 x – log 2 3 = 2 ?
Mathematics
katieb
  • katieb
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
is it log 2x or is it log x to the base 2, and also log 23 or log 3 to base 2
anonymous
  • anonymous
please retype the question
anonymous
  • anonymous
\[\huge \log_2 x - \log_2 3 = 2\]this i guess

Looking for something else?

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

More answers

UMULAS
  • UMULAS
yes the one the np did
UMULAS
  • UMULAS
I mean as usual logs i didd x - = 2 3 But the answer is not 6.
anonymous
  • anonymous
\[\huge \log_2 x = 2+ \log_2 3\]\[\huge 2^{\log_2 x} = 2^{2 + \log_2 3}\]\[\huge x = 2^2 \times 2 ^{\log_2 3}\]\[x = 4 \times 3\]
UMULAS
  • UMULAS
\_/ what how did you get he log and the 2 together?
anonymous
  • anonymous
o.O which one, how to type it or how to do it?
UMULAS
  • UMULAS
how to do it.
anonymous
  • anonymous
\[\huge \log_2 x = 2 + \log_2 3\]then 2^(everything)\[\huge 2^{\log_2 x} = 2^{2+\log_2 3}\]\[\huge 2^{\log_2 x} = x\]and the other side using a^(m+n) = a^m * a^n \[\huge 2^{2+\log_2 3} = 2^2 \times 2 ^{\log_2 3} = 4 \times 3\]

Looking for something else?

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