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mihaellafly
Pe: 2012-11-10T12:19:09+02:00 In: In:

sume de logaritmi

Sa se calculeze suma :

log_{a}x\cdot \sqrt[n]{x^{2}}+log_{a}x^{3}\cdot \sqrt[n]{x^{4}}+ ...+log_{a}x^{2n-1}\cdot \sqrt[n]{x^{2n}}

Sa demonstreze egalitatea pe domeniu de existenta:
\frac{log_{a}\sqrt[n]{x}+og_{a}\sqrt[n]{x^3}+...+log_{a}\sqrt[n]{x^{2n-1}}}{log_{a}\sqrt[n]{x^2}+log_{a}\sqrt[n]{x^4}+...+log_{a}\sqrt[n]{x^{2n}}}=\frac{n}{n+1}

Cat m-am chinuit la aceste doua exercitii si nu le-am gasit o rezolvare corecta ! Va rog frumos sa ma ajutati , daca poate cineva? 🙁

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3 raspunsuri

  1. edy8
    maestru (V)

    mihaellafly wrote:

    Sa demonstreze egalitatea pe domeniu de existenta:
    \frac{log_{a}\sqrt[n]{x}+og_{a}\sqrt[n]{x^3}+...+log_{a}\sqrt[n]{x^{2n-1}}}{log_{a}\sqrt[n]{x^2}+log_{a}\sqrt[n]{x^4}+...+log_{a}\sqrt[n]{x^{2n}}}=\frac{n}{n+1}

    \it{\Large\bl Stim ca:\\\;\\1+3+5+ ... +2n-1 = n^2 (1)\\\;\\2+4+6+ ... +2n = 2(1+2+3+ ... +n) =2\frac{n(n+1)}{2} = n(n+1) (2)\\\;\\log_{a}\sqrt[n]{x}+log_{a}\sqrt[n]{x^3}+...+log_{a}\sqrt[n]{x^{2n-1}} = log_a\sqrt[n]{x\cdot x^3\cdot x^5\cdot ... \cdot x^{2n-1}}=\\\;\\=log_a\sqrt[n]{x^{1+3+5+ ... +2n-1}} \stackrel{(1)}{=} log_a\sqrt[n]{x^{n^2}}= log_a x^{\frac{n^2}{n}} = log_a x^n = nlog_ax (3)\\\;\\log_{a}\sqrt[n]{x^2}+log_{a}\sqrt[n]{x^4}+...+log_{a}\sqrt[n]{x^{2n}} = log_a\sqrt[n]{x^2\cdot x^4\cdot x^6\cdot ... \cdot x^{2n}} =\\\;\\log_a\sqrt[n]{x^{2+4+6+ ... +2n}} \stackrel{(2)}{=} log_a\sqrt[n]{x^{n(n+1)}}= log_a x^{\frac{n(n+1)}{n}} = log_a x^{n +1}= (n+1)log_ax (4)\\\;\\(3), (4) \Rightarrow \frac{log_{a}\sqrt[n]{x}+log_{a}\sqrt[n]{x^3}+...+log_{a}\sqrt[n]{x^{2n-1}}}{log_{a}\sqrt[n]{x^2}+log_{a}\sqrt[n]{x^4}+...+log_{a}\sqrt[n]{x^{2n}}} = \frac{nlog_ax}{(n+1)log_ax} = \frac{n}{n+1}}

  2. gunty
    maestru (V)

    Primul exercitiu:

        \[ \begin{array}{l} \log _a x \cdot \sqrt[n]{{x^2 }} + \log _a x^3 \cdot \sqrt[n]{{x^4 }} + ... + \log _a x^{2n - 1} \cdot \sqrt[n]{{x^{2n} }} = \\ = \log _a \left( {x \cdot x^3 \cdot ... \cdot x^{2n - 1} \cdot \sqrt[n]{{x^2 \cdot x^4 \cdot ... \cdot x^{2n} }}} \right) = \log _a \left( {x^{1 + 3 + ... + (2n - 1)} \cdot \sqrt[n]{{x^{2 + 4 + ... + 2n} }}} \right) = \\ = \log _a \left( {x^{\frac{{1 + (2n - 1)}}{2} \cdot n} \cdot \sqrt[n]{{x^{\frac{{2 + 2n}}{2} \cdot n} }}} \right) = \log _a \left( {x^{n^2 } \cdot \sqrt[n]{{x^{n(n + 1)} }}} \right) = \log _a \left( {x^{n^2 } \cdot x^{n + 1} } \right) = \\ = \log _a x^{n^2 + n + 1} = (n^2 + n + 1) \cdot \log _a x \\ \end{array} \]

  4. mihaellafly

    Va multumesc frumos amandurora!

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