Question

pessoal peço ajuda!! log ½^(x+1)+log½^(x—5)=log½^(2x+3)

Answer 1

Resposta:

$log\frac{1}{2}^(^x^+^1^) + log\frac{1}{2}^(^x^-^5^) = log\frac{1}{2}^(^2^x^+^3^)\\\\(x+1)log\frac{1}{2} + (x-5)log\frac{1}{2} = (2x+3)log\frac{1}{2} \\\\(log\frac{1}{2}.x + log\frac{1}{2}.1) + (log\frac{1}{2}.x + log\frac{1}{2}.-5)=(log\frac{1}{2}.2x + log\frac{1}{2}.3)\\\\(-0,3x - 0,3) + (-0,3x + 0,3.5) =  (-0,3.2x-0,3.3)\\\\(-0,3x -0,3) + (-0,3x + 1,5) = (-0,3.2x - 0,9)\\\\1,2 - 2(0,3.x) = 0,3.2x - 0,9\\\\-2(0,3x) - 0,3.2x = -0,9 - 1,2\\\\-2(0,3x) - 0,3.2x = -2,1$

$-2(0,3x) - 0,3.2x = -2,1\\\\-0,3x - 0,3x = -2,1/4\\\\-0,6x = -0,525\\\\x= -0,525/-0,6\\\\x= 0,875$

A conta da fracionada, pois a base da potência era 10, pois não tem sentido se:

$log\frac{1}{2}$

Tivesse uma base diferente de 10, e o log de:

$log\frac{1}{2}=-0.3010..$

ABRAÇOS