please help

Question

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512

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$Solve for $x$: $$\left(\frac{1}{25}\right)^{x + 2} = 125^{-x}.$$$

Guest Dec 14, 2014

+23254

+5

Best Answer

1/25 = 5^-2 ---> (1/25)^x+2 = (5^-2)^x+2 = 5^-2(x+2)

125 = 5³ ---> (125)^-x = (5³)^-x = 5^-3x

(1/25)^x+2 = (125)^-x ---> 5^-2(x+2)= 5^-3x---> -2(x + 2) = -3x

---> -2x - 4 = -3x

---> x = 4

geno3141 Dec 14, 2014

geno3141 · Answer 1 · 2014-12-14T10:04:55

1/25 = 5^-2 ---> (1/25)^x+2 = (5^-2)^x+2 = 5^-2(x+2)

125 = 5³ ---> (125)^-x = (5³)^-x = 5^-3x

(1/25)^x+2 = (125)^-x ---> 5^-2(x+2)= 5^-3x---> -2(x + 2) = -3x

---> -2x - 4 = -3x

---> x = 4