Find the number of values of x that satisfy t | vedclass

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

Vedclass · Accepted Answer

$2$

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(C) Given equation: $5^{x - 1} + 5 \cdot (0.2)^{x - 2} = 26$ Since $0.2 = \frac{1}{5} = 5^{-1}$,we have: $5^{x - 1} + 5 \cdot (5^{-1})^{x - 2} = 26$ $5^{x - 1} + 5 \cdot 5^{-(x - 2)} = 26$ $5^{x - 1} + 5 \cdot 5^{2 - x} = 26$ $5^{x - 1} + 5^{3 - x} = 26$ $5^{x - 1} + \frac{5^3}{5^x} = 26$ $5^{x - 1} + \frac{125}{5^x} = 26$ Let $y = 5^{x - 1}$. Then $5^x = 5y$. $y + \frac{125}{5y} = 26$ $y + \frac{25}{y} = 26$ $y^2 - 26y + 25 = 0$ $(y - 25)(y - 1) = 0$ So,$y = 25$ or $y = 1$. If $5^{x - 1} = 25 = 5^2$,then $x - 1 = 2$,so $x = 3$. If $5^{x - 1} = 1 = 5^0$,then $x - 1 = 0$,so $x = 1$. There are $2$ values of $x$ ($1$ and $3$) that satisfy the equation.

Find the number of values of $x$ that satisfy the equation $5^{x - 1} + 5 \cdot (0.2)^{x - 2} = 26$.

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