208/VB02. problem / Math - Solved Math Problems

MR-208 / 1076a. problem

$5^{x} - 5^{3 - x} = 20$

$5^{x} - 5^{3 - x} = 20$

5^{x} - 5^{3 - x} = 20

5^{x} - \frac{5^{3}}{5^{x}} = 20

5^{x} = t \Rightarrow t - \frac{5^{3}}{t} = 20

t - \frac{125}{t} = 20 \cdot t

t^{2} - 125 = 20 t

t^{2} - 20 t - 125 = 0

t^{2} - 20 t - 125 = 0 \Rightarrow a = 1; b = - 20; c = - 125

t_{1, 2} = \frac{20 \pm \sqrt{{(- 20)}^{2} - 4 \cdot 1 (- 125)}}{2} = \frac{20 \pm \sqrt{400 + 500}}{2} = \frac{20 \pm \sqrt{900}}{2} = \frac{20 \pm 30}{2}

t_{1} = 25; t_{2} = - 5

t_{1} = 25 \Rightarrow 5^{x_{1}} = 25 \Rightarrow 5^{x_{1}} = 5^{2} \Rightarrow x_{1} = 2

t_{2} = - 5 \Rightarrow 5^{x_{2}} > 0 \Rightarrow 5^{x_{2}} \neq - 5

$x = 2$

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