exponential equation

\hspace{-16}$Solve for $\mathbf{x}$ in \\\\ $\mathbf{2^x+3^x+4^x+5^x+1=2e^x+3.7^x}$

x=0

3 Answers

262
Aditya Bhutra ·

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

71
Vivek @ Born this Way ·

Wah!

1708
man111 singh ·

Thanks Aditiya very Nice Man

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