- #1
juantheron
- 247
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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.
. 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.