Question

${81}^{x +1} = ({ \frac{1}{3} })^{x2}$
Equação exponencial. ​

Answer 1

Resposta:

81^(x+1) =(1/3)^x²

(3^4)^(x+1) =(3^(-1))^x²

3^(4x+4)= 3^(-x²)

4x+4 =-x2

x²+4x+4 =0

(x+2)²=0

x+2=0  ==>x=-2

Answer 2

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{81^{x + 1} = \left(\dfrac{1}{3}\right)^{x^2}}$

$\mathsf{(3^4)^{x + 1} = (3^{-1})^{x^2}}$

$\mathsf{3^{4x + 4} = 3^{-x^2}}$

$\mathsf{4x + 4 = -x^2}$

$\mathsf{x^2 + 4x + 4 = 0}$

$\mathsf{(x + 2)^2 = 0}$

$\mathsf{x + 2 = 0}$

$\boxed{\boxed{\mathsf{x = -2}}}$