Question

log [8x+2] - log [x-4] = 1 - log 2

Answer 1

Respuesta:

$x=-\frac{8}{15}$

Explicación paso a paso:

$log(8x+2)-log(x-4)=1-log2\\log(\frac{8x+2}{x-4})=1-log2\\ log(\frac{8x+2}{x-4} )+log2=1\\\\log(2(\frac{8x+2}{x-4}))=1\\ log(\frac{16x+4}{x-4})=1\\(x-4)^{1} =16x+4\\x-4=16x+4\\16x-x=-4-4\\15x=-8\\x=-\frac{8}{15}$