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jhonyy9

  • one year ago

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 ?

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  1. jhonyy9
    • one year ago
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    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

  2. jhonyy9
    • one year ago
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    - so than i remember right again the accepted correct root was x= 6^(1/3) - why ?

  3. anonymous
    • one year ago
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    \(2^{3^4}\) or \((2^3)^4\)?

  4. anonymous
    • one year ago
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    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 \(3>2\implies 3^4>2^4\implies A=2^{3^4}>2^{2^4}=4^{2^3}=B\)

  5. anonymous
    • one year ago
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    $$(6x)^6=6^{2^3}\\6^6 x^6=6^8\\x^6=6^2\\x^3=\pm6\\x=\pm\sqrt[3]{6}$$

  6. dan815
    • one year ago
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    |dw:1436490463980:dw|

  7. dan815
    • one year ago
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    |dw:1436490539513:dw|

  8. dan815
    • one year ago
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    perhaps trying to find a generalization for this is more intriguing

  9. dan815
    • one year ago
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    like does this hold for all consecutive numbers

  10. dan815
    • one year ago
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    |dw:1436490662853:dw|

  11. anonymous
    • one year ago
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    the first one is exactly what I wrote; I don't understand why you just drew it out...

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