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OVIDIU1971
Pe: 2012-10-30T19:07:55+02:00 In: In:

problema puteri

care numar este mai mare 15 la puterea 26 sau 31 la puterea 19?

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6 raspunsuri

  1. Minecraft
    maestru (V)

        \[\left\{ {\left. \begin{array}{l} {15^{26}} = {15^{7 + 19}} = {15^7}*{15^{19}}\\ {\rm{3}}{{\rm{1}}^{19}}\rangle {30^{19}} = {2^{19}}*{15^{19}} \end{array} \right\}} \right. \Rightarrow {\rm{ }}{2^{19}}si{\rm{ }}{15^7} \Rightarrow {{\rm{(15 + 1)}}^{15}}\rangle {15^7} \Rightarrow {\rm{3}}{{\rm{1}}^{19}}\rangle {30^{19}}\rangle {15^{26}}\]

  2. ali
    maestru (V)

    31^19 niciodată nu-i mai mare ca 15^26 😀
    Demonstratie:
    a = {15^{26}}\\ b = {31^{19}} < {30^{20}} = {15^{20}} \times {2^{20}} \to \left( {{\rm{conform inegalitatea 1}}} \right)\\ {\rm{Presupunem ca }}{30^{20}} > {15^{26}} \Leftrightarrow {15^{20}} \times {2^{20}} > {15^{26}} \Leftrightarrow {2^{20}} > {15^6} \Leftrightarrow {16^5} > {15^6} \to {\rm{fals}}\\ \to \left( {{\rm{conform inegalitatea 1}}} \right) \Rightarrow {31^{19}} < {30^{20}} < {15^{26}}\\
    \begin{array}{l} {\rm{Acum demonstrez}}:{k^{k + 1}} > {\left( {k + 1} \right)^k}, \forall k \ge 3 \to \overline {\underline {\left| {{\rm{inegalitatea1}}} \right|} } \\ k \times {k^k} > {\left( {k + 1} \right)^k} \Leftrightarrow k > {\left( {\frac{{k + 1}}{k}} \right)^k} \Leftrightarrow k > {\left( {1 + \frac{1}{k}} \right)^k} \to {\rm{inegalitatea 2}}\\ {\rm{Dar}}:{\left( {1 + \frac{1}{k}} \right)^k} < 3, \Rightarrow {\rm{inegalitatea 2}} = {\rm{adevarata}} \Leftrightarrow {\rm{k}} \ge {\rm{3}}. \end{array}
    Finish …..

  4. edy8
    maestru (V)

    ali wrote:
    {\rm{Dar}}:{\left( {1 + \frac{1}{k}} \right)^n} < 3, \forall n \in {N^*} \Rightarrow {\rm{inegalitatea 2}} = {\rm{adevarata}} \Leftrightarrow {\rm{k}} \ge {\rm{3}}.
    .

    …?!

  5. ali
    maestru (V)

    Mai exact, care-i întrebarea ?
    De ce asta:

    {\left( {1 + \frac{1}{k}} \right)^n} < 3


    Sau :

    k \ge 3

  6. edy8
    maestru (V)

    ali wrote:
    {\rm{Dar}}:{\left( {1 + \frac{1}{k}} \right)^n} < 3, \forall n \in {N^*} \Rightarrow {\rm{inegalitatea 2}} = {\rm{adevarata}} \Leftrightarrow {\rm{k}} \ge {\rm{3}}.
    .

    …?!

    \it{\bl k=3, n=4\\\;\\(1+\frac{1}{3})^4 > 3}

  8. ali
    maestru (V)

    Este o greseala acolo la fractie nu este k ci n sau ma rog în loc de n la putere este k.
    Acea afirmatie este bazata pe: {\left( {1 + \frac{1}{n}} \right)^n} \to e
    Am gresit la notatii de la început. 🙁 voi corecta !

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