Question

como se resolve :
81^x-1 = ( 1/3) ^x^2

Answer 1

Explicação passo-a-passo:

$81 {}^{x - 1} = ( \dfrac{1}{3} ) {}^{x {}^{2} }$

$3 {}^{4x - 4} = 3 {}^{ - x {}^{2} }$

$4x - 4 = - x {}^{2}$

$4x - 4 + x {}^{2} = 0$

$x {}^{2} + 4x - 4 = 0$

$x = \dfrac{ - 4 \dfrac{ + }{} \sqrt{4 {}^{2} - 4 \times 1 \times - 4} }{2 \times 1}$

$x = \dfrac{ - 4 \dfrac{ + }{} \sqrt{16 + 16} }{2}$

$x = \dfrac{ - 4 \dfrac{ + }{} \sqrt{32} }{2}$

$x = \dfrac{ - 4 \dfrac{ + }{} 2 {}^{2} \sqrt{2} }{2}$

$x = \dfrac{ - 4 \dfrac{ + }{}4 \sqrt{2} }{2}$

$x {}^{1} = \dfrac{ - 4 + 4 \sqrt{2} }{2}$

$x {}^{1} = - 2 + 2 \sqrt{2}$

$x {}^{2} = \dfrac{ - 4 - 4 \sqrt{2} }{2}$

$x {}^{2} = - 2 - 2 \sqrt{2}$