If 52+4+6+⋯+2x=(0.04)−28 and (x>0), then x=

52+4+6+⋯+2x=(0.04)−2852(1+2+3+⋯+x)=(125)−2852×x(x+1)2=(125)−28=(5−2)−285x(x+1)=556x(x+1)=56x2+x−56=0x2+8x−7x−56=0(x+8)(x−7)=0x=-8 (not possible), x=7possible

If 52+4+6+⋯+2x=(0.04)−28 and (x>0), then x=

52+4+6+⋯+2x=(0.04)−2852(1+2+3+⋯+x)=(125)−2852×x(x+1)2=(125)−28=(5−2)−285x(x+1)=556x(x+1)=56x2+x−56=0x2+8x−7x−56=0(x+8)(x−7)=0x=-8 (not possible), x=7possible

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If $5^{2 + 4 + 6 + \dots + 2 x} = {(0.04)}^{- 28}$ and $(x > 0)$ , then $x =$

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Detailed Solution

$\begin{array}{l} 5^{2 + 4 + 6 + \dots + 2 x} = {(0.04)}^{- 28} \\ 5^{2 (1 + 2 + 3 + \dots + x)} = {(\frac{1}{25})}^{- 28} \\ 5^{2 \times \frac{x (x + 1)}{2}} = {(\frac{1}{25})}^{- 28} = {(5^{- 2})}^{- 28} \\ 5^{x (x + 1)} = 5^{56} \\ x (x + 1) = 56 \\ x^{2} + x - 56 = 0 \\ x^{2} + 8 x - 7 x - 56 = 0 \\ (x + 8) (x - 7) = 0 \end{array}$