Question

Resolva as seguintes equações exponenciais :/   $10^{3X} =100.000/$                         $9^{X+1}= \sqrt[3]{3}$/                  $8^{X}=0,125$ /                                                      $8^{2X+1} =0$

Answer 1

$10^{3x}=100.000 \\ 10^{3x} = 10^{5} \\ 3x = 5 \\ x = \frac{5}{3} \\ \\ 9^{x+1}= \sqrt[3]{3} \\ 3^{2(x+1)}= 3^{ \frac{1}{3} } \\ (2x+2) = \frac{1}{3} \\ 6x + 6 = 1 \\ 6x = -5 \\ x = - \frac{5}{6} \\ \\ 8^{x} = 0,125 \\ 8^{x} = \frac{1}{8} \\ 8^{x} = 8^{-1} \\ x = -1 \\ \\ 8^{2x+1} = 0,125 \\ 8^{2x+1} = \frac{1}{8} \\ 8^{2x+1} = 8^{-1} 2x + 1 = -1 \\ 2x = -1 -1 \\ 2x = -2 \\ x = -1$


Espero ter ajudado.