Question

Equação exponencial, qual o resultado: $\sqrt 5^{x-2} . \sqrt[x]25^{2x-5} - \sqrt[2x] 5^{3x+2} =0$

Answer 1

(0,5)^2x = 2^1-3x
(1/2)^2x = 2^1-3x
(2-¹)^2x = 2^1-3x
(2)^-2x = 2^1-3x
-2x = 1-3x
-2x+3x = 1
x = 1

Answer 2

Podemos escrever essa equação como:

$5^{\dfrac{x-2}{2}}\times 5^{\dfrac{2(2x-5)}{x}}=5^{\dfrac{3x+2}{2x}}$

$5^{\dfrac{x-2}{2}+\dfrac{2(2x-5)}{x}}=5^{\dfrac{3x+2}{2x}}$

$\dfrac{x-2}{2}+\dfrac{4x-10}{x}=\dfrac{3x+2}{2x}$

$\dfrac{x-2}{2}+\dfrac{8x-20}{2x}-\dfrac{3x+2}{2x}=0$

$\dfrac{x^2-2x+8x-20-3x-2}{2x}=0$

$x^2+3x-22=0$

Por Báskara:

$\Delta = 3^2-4\times 1\times (-22)=97$

$x=\dfrac{-3\pm \sqrt{97}}{2}$


Espero ter ajudado!