Question

тренажер по теме показательные уравнения 10 класс подготовка к ЕГЭ

Answer 1

$1.;(0,5)^x=frac1{64}\left(frac12right)^x=left(frac12right)^6\x=6\\2.;sqrt[3]{128}=4^{2x}\128^{frac13}=4^{2x}\left(2^7right)^{frac13}=(2^2)^{2x}\2^{frac73}=2^{4x}\frac73=4x\x=frac7{12}\\3.;3^{x^2-4x-0,5}=81sqrt3\3^{x^2+4x-0,5}=3^4cdot3^{frac12}\3^{x^2+4x-0,5}=3^{4,5}\x^2+4x-0,5=4,5\x^2+4x-5=0\D=16+4cdot5=36\x_{1,2}=frac{-4pm6}2\x_1=-5,;x_2=1$

$4.;left(frac37right)^{3x-7}=left(frac73right)^{7x-3}\left(frac37right)^{3x-7}=left(frac37right)^{3-7x}\3x-7=3-7x\3x+7x = 3+7\10x=10\x=1\\5.;left(frac23right)^xcdotleft(frac98right)^x=frac{27}{64}\left(frac23cdotfrac98right)^x=frac{3^3}{4^3}\left(frac34right)^x=left(frac34right)^3\x=3\\6.;(0,4)^{x-1}=(6,25)^{6x-5}\(2,5)^{1-x}=(2,5^2)^{6x-5}\(2,5)^{1-x}=(2,5)^{12x-10}\1-x=12x-10\-x-12x=-10-1\-13x=-11\x=frac{11}{13}$

$7.;3^xcdotleft(frac13right)^{x-3}=left(frac1{27}right)^x\3^xcdot3^{3-x}=3^{-3x}\3^3=3^{-3x}\3=-3x\x=-1\\8.;5^{2x+1}-3cdot5^{2x-1}=550\5cdot5^{2x}-frac35cdot5^{2x}=550\frac{22}5cdot5^{2x}=550\5^{2x}=550:frac{22}5=550cdotfrac5{22}=125\5^{2x}=5^3\2x=3\x=frac23\\9.;sqrt{3^x}cdot5^{frac{x}2}=225\3^{frac x2}cdot5^{frac x2}=225\(3cdot5)^{frac x2}=225\15^{frac x2}=15^2\frac x2=2\x=4$

$10.;2^{sqrt x+2}-2^{sqrt x+1}=12+2^{sqrt x-1}\2^{sqrt x+2}-2^{sqrt x+1}-2^{sqrt x-1}=12\4cdot2^{sqrt x}-2cdot2^{sqrt x}-frac12cdot2^{sqrt x}=12\frac32cdot2^{sqrt x}=12\2^{sqrt x}=8\2^{sqrt x}=2^3\sqrt x=3\x=9\\11.;5^{2x}-2cdot5^x-15=0\5^x=t,;5^{2x}=t^2,;t textgreater 0\t^2-2t-15=0\D=4+4cdot15=64\t_{1,2}=frac{2pm8}2\t_1=5\t_2=-3;-;He;nogx.\5^x=5\x=1$

$12.;4^{x^2+2}-9cdot2^{x^2+2}+8=0\2^{2x^2+4}-9cdot2^{x^2+2}+8=0\2^{x^2+2}=t,;2^{2x^2+4}=t^2,;t textgreater 0\t^2-9t+8=0\D=81-32=49\t_{1,2}=frac{9pm7}2\t_1=1,;t_2=8\2^{x^2+2}=1\x^2+2=0;-;pew.;HET\2^{x^2+2}=8\x^2+2=3\x^2=1\x=1$

$13.;4^{3x^2+x}-8=2cdot8^{x^2+frac x3}\(2^2)^{3x^2+x}-8=2cdot(2^3)^{x^2+frac x3}\2^{6x^2+2x}-8=2cdot2^{3x^2+x}\2^{6x^2+2x}-2cdot2^{3x^2+x}-8=0\2^{3x^2+x}=t,;2^{6x^2+2x}=t^2,;t textgreater 0\t^2-2t-8=0\D=4+4cdot8=36\t_{1,2}=frac{2pm6}2\t_1=4\t_2=-3;-;He;nogx.\2^{3x^2+2}=4\2^{3x^2+2}=2^2\3x^2+2=2\3x^2=0\x=0\\14.;4^{x-1}+4^x+4^{x+1}=84\frac14cdot4^x+4^x+4cdot4^x=84\frac{21}4cdot4^x=84\4^x=16\4^x=4^2\x=2$