Question

ajuda para resolver equação exponencial:

$64^{2x+1} . 8^{x+3} / 4096 = \frac{1}{8} x^$

Answer 1

64^2x+1 .8^x+3 = (1)^x
         4096              8

(8^2)^2x+1 .8^x+3 = 4096.(8^)-1^x

8^4x+2 .8^x+3 = 8^4.8^-x ==> 8^(4x+2+x+3) = 8^(4-x)

5x+5 = 4-x ==> 5x+x = 4-5 ==> 4x = - 1 ==> x = - 1/4

Answer 2

$64 = 2^{6} \\ 8 = 2^{3} \\ 4096 = 2^{12}$
$\frac{1}{8} = \frac{1}{ 2^{3} } \\ 1 = 2^{0}$
Substituindo esses valores na fómula, temos:
$\frac{(2^{6})^{2x+1}*(2^{3})^{x+3} }{2^{12}} = \frac{2^{0}}{(2^{3})^{x}} \\ 2^{12x+6}*2^{3x+9}*2^{-12} = 2^{0}*2^{-3x} \\ 2^{12x+6+3x+9-12} = 2^{0-3x} \\ 2^{15x+3} = 2^{-3x} \\ 15x+3 = -3x \\ 15x + 3x = -3 \\ 18x =-3 \\ x = \frac{-3}{18} = \frac{-1}{6}$