Question

помогите пожайлуста решить 2,7 и 2,12​

Answer 1

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Answer 2

2.7

1)

$3^x = \frac{1}{9}\\3^x = \frac{1}{3^2}\\3^x = 3^{-2}\\x = -2$

2)

$5^x = \sqrt{5}\\5^x = 5^{\frac{1}{2}}\\x = \frac{1}{2}$

3)

$7^x = \sqrt[3]{49} \\7^x = \sqrt[3]{7^2} \\7^x = 7^{\frac{2}{3}}\\x = \frac{2}{3}$

4)

$(\frac{2}{3})^x = 1,5\\(\frac{2}{3})^x = \frac{3}{2}\\(\frac{2}{3})^x = (\frac{2}{3})^{-1} \\x = -1$

2.12

1)

$7^{x^2-x-2} = 1\\7^{x^2-x-2} = 7^0\\x^2-x-2 = 0\\D = (-1)^2 - 4*1*(-2) = 1 + 8 = 9\\x_{1} = \frac{1+3}{2} = \frac{4}{2} = 2\\x_{2} = \frac{1-3}{2} = \frac{-2}{2} = -1$

2)

$4^{x^2+2x} = 64\\4^{x^2+2x} = 4^3\\x^2+2x-3 = 0\\D = 4 + 12 = 16\\x_{1} = \frac{-2+4}{2} = \frac{2}{2} = 1\\x_{2} = \frac{-2-4}{2} = \frac{-6}{2} = -3$

3)

$(\frac{1}{8})^{x^2-2x} = (\frac{1}{8})^{2-3x}\\x^2 - 2x = 2-3x \\x^2 + x - 2 =0;\\D = 1 +8 = 9\\x_{1} = \frac{-1+3}{2} = \frac{2}{2} = 1\\x_{2} = -2$

4)

$2^{x^2} = 8\\2^{x^2} = 2^3\\x^2 = 3\\x_{1} = \sqrt{3}\\x_{2} = - \sqrt{3}$