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Exponential equation

  1. Nov 5, 2013 #1
    No. of Real solution of the equation ##2^x\cdot \ln (2)+3^x\cdot \ln (3)+4^x\cdot \ln (4) = 2x##

    . The attempt at a solution

    Let ##f(x) = 2^x\cdot \ln (2)+3^x\cdot \ln (3)+4^x\cdot \ln (4) -2x##

    Here solution must be exists for ##x>0##, because for ##x\leq 0##, L.H.S>0 while R.H.S <0

    ##f^{'}(x) = 2^x\cdot \left(\ln 2\right)^2+3^x\cdot \left(\ln 3\right)^2+4^x\cdot \left(\ln 4\right)^2 - 2##

    and ##f^{''}(x) = 2^x\cdot \left(\ln 2\right)^3+3^x\cdot \left(\ln 3\right)^3+4^x\cdot \left(\ln 4\right)^3 >0\forall x\in \mathbb{R}##

    Means function ##f(x)## is Concave - Upward and ##f^{'}(x)## is Strictly Increasing function

    and when ##x\rightarrow +\infty, f^{'}(x)\rightarrow \infty ## and when ##x\rightarrow -\infty, f^{'}(x)\rightarrow -\infty ##

    So using LMVT, function ##f^{'}(x) = 0## has exactly one real root.

    But I did not understand How can i calculate root of ##f(x) = 0##

    Help please

    Thanks in Advance.
     
  2. jcsd
  3. Nov 5, 2013 #2

    Mentallic

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    Are you sure about that?

    There isn't a real root, but even if there were, this analysis won't give you the value of the root(s).

    Can you show that f'(x) crosses the x-axis only once? And also, since you've already shown that f''(x) > 0 for all x hence f(x) is concave upwards, what can you say about f(x) where x is the turning point (where f'(x)=0)?

    If you can't quite understand this, picture a general parabola that's concave upwards. It's essentially the same problem.
     
  4. Nov 5, 2013 #3

    Ray Vickson

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    The function g(x) = ln(2)*2^x + ln(3)*3^x + ln(4)*4^x - 2*x is strictly convex on the real line (what you call concave up --- an archaic terminology no longer used by optimization people), so there are either 0, 1 or 2 real solutions of g(x) = 0. What you checked is not sufficient to establish the existence of a real root; can you really say for sure that there must be a root?
     
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