Question

Rozwiaz równania wykładnicze

Answer 1

Verified answer

[tex]9=(\frac{1}{3})^{-2}[/tex]

[tex](\frac{1}{3} )^{-2*(2x-1)}=(\frac{1}{3}) ^{5x-12}[/tex]

[tex](\frac{1}{3} )^{-4x+2}=(\frac{1}{3}) ^{5x-12}[/tex]

[tex]-4x+2=5x-12\\-4x-5x=-12-2\\-9x=-14 |:(-9)\\x=\frac{14}{9} = 1\frac{5}{9}[/tex]

[tex](\frac{2}{5})^{x^{2} -3x}=(\frac{2}{5})^{-2x+6}[/tex]

[tex]x^{2} -3x=-2x+6\\x^{2} -3x+2x-6=0\\x^{2} -x-6=0[/tex]

[tex]x^{2} -x-6=0\\(x+2)(x-3)=0\\x_{1} =-2\\x_{2} =3[/tex]

Answer 2

Odpowiedź:

[tex]9^{2 x - 1} = ( \frac{1}{3})^{5 x - 12}\\ (3^2)^{2 x - 1} = ( 3^{-1})^{5 x - 12}\\3^{4 x - 2} = 3^{- 5 x + 12 }\\4 x - 2 = - 5 x + 12\\9 x = 14 / : 9\\x = \frac{14}{9} = 1 \frac{5}{9}[/tex]

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[tex](\frac{2}{5} )^{x^{2} -3 x} = ( \frac{2}{5} )^{-2 x + 6}\\x^{2} - 3 x = -2 x + 6\\x^{2} - x - 6 = 0\\( x + 2)*( x - 3) = 0\\x + 2 = 0[/tex]  lub        x - 3 = 0

x =  - 2      lub     x = 3

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