Question

Ajutor, am nevoie urgent!!!

Answer 1

 

$\displaystyle\\a)\\125^x=\frac{1}{5}\\\\\Big(5^3\Big)^x=5^{-1}\\\\ 5^{3x}=5^{-1}\\\\3x=-1\\\\\boxed{x=-\frac{1}{3}}\\\\\\b)\\\frac{~1~}{2^x}=4\\\\ 2^{-x}=2^2\\-x=2\\\boxed{x=-2}\\\\c)\\\left(\frac{1}{2}\right)^x=2^{x-2}\\\\2^{-x}=2^{x-2}\\-x=x-2\\-x-x=-2\\-2x=-2\\2x=2\\\\x=\frac{2}{2}\\\\ \boxed{x=1}$

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$\displaystyle\\d)\\4^{x+2}=2^{x^2-4}\\\\\Big(2^2\Big)^{x+2}=2^{x^2-4}\\\\\2^{2(x+2)}=2^{x^2-4}\\2(x+2)=x^2-4\\2x+4=x^2-4\\x^2-2x-4-4=0\\x^2-2x-8=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{2\pm\sqrt{4+32}}{2}=\\\\=\frac{2\pm\sqrt{36}}{2}=\frac{2\pm6}{2}=1\pm3\\\\x_1=1+3\\\boxed{x_1=4}\\\\x_2=1-3\\\boxed{x_2=-2}$

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$\displaystyle\\e)\\2^{x^2-4x}=\frac{1}{8}\\\\2^{x^2-4x}=\frac{1}{2^3}\\\\2^{x^2-4x}=2^{-3}\\x^2-4x=-3\\x^2-4x+3=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{4\pm\sqrt{16-12}}{2}=\\\\=\frac{4\pm\sqrt{4}}{2}=\frac{4\pm2}{2}=2\pm1\\\\x_1=2+1\\\boxed{x_1=3}\\\\x_2=2-1\\\boxed{x_2=1}\\\\\\f)\\\\\frac{2^x}{3^x}=\frac{3}{2}\\\\\left(\frac{2}{3}\right)^x=\left(\frac{2}{3}\right)^{-1}\\\\\boxed{x=-1}$