11881
punti
Livello 2
► C.E.
C.E. 0 < x < 1
► Risoluzione
$ log ((1-x)(2-x)) = log (x(x+5)) \; ⇒ \; (1-x)(2-x) = x(x+5) \; ⇒ \; 2-8x = 0 \; ⇒ \; x = \frac{1}{4} $
21742
punti
Livello 6
=======================================================
$\small log(1-x)+log(2-x) = log(x) +log(x+5)$
$\small log_{10}(1-x)+log_{10}(2-x) = log_{10}(x) +log_{10}(x+5)$
$\small log_{10}[(1-x)(2-x)]=log_{10}[x(x+5)]$
$\small log_{10}(2-x-2x+x^2)=log_{10}(x^2+5x)$
$\small log_{10}(x^2-3x+2)=log_{10}(x^2+5x)$
$\small x^2-3x+2 = x^2+5x$
$\small x^2-3x-x^2-5x = -2$
$\small \cancel{x^2}-8x-\cancel{x^2} = -2$
$\small -8x = -2$
$\small \dfrac{-\cancel8x}{-\cancel8} = \dfrac{-\cancel2^1}{-\cancel8_4}$
$\small x = \dfrac{1}{4}$
68277
punti
Genius I