Question

resolva,em r, as seguintes equações exponenciais (1/9)^x²-1.27^1-x=3^2x+7​

Answer 1

${ (\frac{1}{9}) }^{ {x}^{2} - 1 } \times {(27)}^{1 - x} = {3}^{2x + 7} \\ {3}^{ - 2 {x}^{2} + 2} \times {3}^{3 - 3x} = {3}^{2x + 7} \\ {3}^{ - 2 {x}^{2} - 3x + 5 } = {3}^{2x + 7} \\ \\ 2 {x}^{2} + 5x + 2 = 0 \\ \\ x = \frac{ - 5 \frac{ + }{} \sqrt{25 - 16} }{4} \\ x = \frac{ - 5 \frac{ + }{}3 }{4} \\ \\ x = - \frac{1}{2} \: ou \: x = - 2$

Answer 2

Resposta:

S = {-2, -1/2}

Explicação passo-a-passo:

$(\frac{1}{3})^{2}^{(x^{2}-1 }^{)}.3^{3}^{(1-x}^{)}=3^{2x+7}\\\\3^{2(1-x^{2}).3^{3^{(1-x)}}}=3^{2x+7}\\\\3^{2(1-x^{2})+3(1-x) }=3^{2x+7}\\\\2-2x^2+3-3x=2x+7\\\\-2x^2-5x-2=0\\\\2x^2+5x+2=0\\\\Delta=5^2-4.2.1\\\\Delta=25-16\\\\Delta=9\\\\x=\frac{-5-3}{2.2}=-2\\ou\\x=\frac{-5+3}{2.2}=-\frac{1}{2}$