\hspace{-16}$Solve for $\mathbf{x}$ in \\\\ $\mathbf{2^x+3^x+4^x+5^x+1=2e^x+3.7^x}$
x=0
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3 Answers
Aditya Bhutra
·2012-03-08 22:54:57
the eqn. can be rearranged as ,
(7x-5x) + (7x-4x)+ (7x-3x) + (ex-2x) + (ex-1) =0
now each of the terms inside the brackets is increasing
hence LHS is increasing
thus only one trivial soln , x=0