jhonyy9
  • jhonyy9
dear Champions ! give me please an explication - thank you - what is larger : A = 2^3^4 or B = 4^2^3 ? - how can i deciding that in case of A or in case of B the top exponent is an exponent of the bottom number ,in case A of 2 or in case B of 4 , or are exponent of the first exponent,in case A ,of 3 or in case B of 2 ?
Mathematics
schrodinger
  • schrodinger
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

jhonyy9
  • jhonyy9
Dear Champions this was what have surprised me in Brilliant.org ,math. forum when i have wrote for above exercise that are equal A and B - so this was a correct accepted answer but after this i have got a second exercise what was : (6x)^6 = 6^2^3 and for this equation i have wrote like firstly very easy root x=1 - so this was what have surprised me when have got the answer that x=1 is wrong so that not is root of this equation - i dont understand it why ? can somebody explaining me please ? thank you very much
jhonyy9
  • jhonyy9
- so than i remember right again the accepted correct root was x= 6^(1/3) - why ?
anonymous
  • anonymous
\(2^{3^4}\) or \((2^3)^4\)?

Looking for something else?

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

More answers

anonymous
  • anonymous
if you mean \(2^{3^4}\) then consider \(4^{2^3}=2^{2\cdot2^3}=2^{2^4}\) so clearly \(2^{3^4}>2^{2^4}\) since \(3>2\implies 3^4>2^4\implies A=2^{3^4}>2^{2^4}=4^{2^3}=B\)
anonymous
  • anonymous
$$(6x)^6=6^{2^3}\\6^6 x^6=6^8\\x^6=6^2\\x^3=\pm6\\x=\pm\sqrt[3]{6}$$
dan815
  • dan815
|dw:1436490463980:dw|
dan815
  • dan815
|dw:1436490539513:dw|
dan815
  • dan815
perhaps trying to find a generalization for this is more intriguing
dan815
  • dan815
like does this hold for all consecutive numbers
dan815
  • dan815
|dw:1436490662853:dw|
anonymous
  • anonymous
the first one is exactly what I wrote; I don't understand why you just drew it out...

Looking for something else?

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