Question

resolva em R as seguintes equações exponenciais​

Answer 1

$0,2^{x+1}=\sqrt{125}$

$(\frac{1}{5})^{x+1}=\sqrt{25\cdot 5}$

$(5^{-1})^{x+1}=5\sqrt{5}$

$5^{-x-1}=5\cdot 5^{1/2}$

$5^{-x-1}=5^{1+1/2}$

$-x-1=1+\frac{1}{2}$

$-x=1+\frac{1}{2}+1$

$-x=\frac{2}{2}+\frac{1}{2}+\frac{2}{2}$

$-x=\frac{5}{2}$

$x=-\frac{5}{2}$

Answer 2

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{0,2^{x + 1} = \sqrt{125}}$

$\mathsf{\left(\dfrac{1}{5}\right)^{x + 1} = 5\sqrt{5}}$

$\mathsf{5^{-x - 1} = 5^1.5^{\frac{1}{2}}}$

$\mathsf{\not5^{-x - 1} = \not5^{\frac{3}{2}}}$

$\mathsf{-x - 1 = \dfrac{3}{2}}$

$\mathsf{-x = \dfrac{3}{2} + 1}$

$\boxed{\boxed{\mathsf{x = -\dfrac{5}{2}}}}$