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**. 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.