. asdf: a) a^x*a^y = a^x+y b) (a^x )^y =a^x*y 32^{(x+5)/(x−7)} = 1/4 * 128 ^{(x+17)/(x−3)} // wzór b) 2^{5 * (x+5)/(x−7)} = 2⁻² * 2^{7*(x+17)/(x−3)} // teraz dla prawej strony stosuje a) 2^{5 * (x+5)/(x−7)} = 2^{−2 + 7*(x+17)/(x−3)} (5x+25)/(x−7) = −2 + (7x+17*7)/(x−3) 5x+25 / x−7 = −2 + (7x + 119)/(x−3) 5x+25 / x−7 = (−2x + 6 + 7x + 119)/(x−3) 5x + 25 / x−7 = 5x + 125 / x−3 5(x+5)/(x−7) = 5(x+25)/(x−3) (x+5)/(x−7) = (x+25)/(x−3) (x−7)(x+25) = (x+5)(x−3) x² + 25x − 7x − 175 = x² + 2x − 15 18x − 175 = 2x − 15 16x = 160 x = 10

bezendu: ok mam

asdf: (0.5)^x2 * 2^2x+2 = 64⁻¹ 2^{−x2 + 2x + 2} = 2⁻⁶ −x² + 2x + 8 = 0

asdf: ⁿ√a = a^1/n

ax
	= ax−y
ay

bezendu:

	√2
0,125*42x−3=(		)−x
	3

2⁻³*2^4x−6=2^2,5x 2^−3+4x−6=2^2,5x 4x−9=2,5x 1,5x=9 x=6

bezendu:

	3		7
(		)3x−7=(		)7x−3
	7		3

	7		7
(		)−3x+7=(		)7x−3
	3		3

−3x+7=7x−3 −10x=−10 x=1