If 5^{x-1} + 5 cdot (0.2)^{x-2} = 26,then the | vedclass

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(B, C) Given equation: $5^{x-1} + 5 \cdot (0.2)^{x-2} = 26$Since $0.2 = \frac{1}{5} = 5^{-1}$,we can rewrite the equation as:$5^{x-1} + 5 \cdot (5^{-1

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(B, C) Given equation: $5^{x-1} + 5 \cdot (0.2)^{x-2} = 26$Since $0.2 = \frac{1}{5} = 5^{-1}$,we can rewrite the equation as:$5^{x-1} + 5 \cdot (5^{-1})^{x-2} = 26$$5^{x-1} + 5 \cdot 5^{-x+2} = 26$$5^{x-1} + 5^{1-x+2} = 26$$5^{x-1} + 5^{3-x} = 26$Let $y = 5^{x-1}$. Then $5^{3-x} = 5^{2-(x-1)} = \frac{5^2}{5^{x-1}} = \frac{25}{y}$.The equation becomes: $y + \frac{25}{y} = 26$$y^2 - 26y + 25 = 0$$(y-25)(y-1) = 0$So,$y = 25$ or $y = 1$.Case $1$: $5^{x-1} = 25 = 5^2 \implies x-1 = 2 \implies x = 3$.Case $2$: $5^{x-1} = 1 = 5^0 \implies x-1 = 0 \implies x = 1$.Thus,$x$ can be $1$ or $3$.

If $5^{x-1} + 5 \cdot (0.2)^{x-2} = 26$,then the value of $x$ can be:

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