Question

Log(2x+1)-log(2x-1)=2log3-3log2

Answer 1

$\log _{10}\left(2x+1\right)-\log _{10}\left(2x-1\right)=2\log _{10}\left(3\right)-3\log _{10}\left(2\right)$
Se suma log₁₀ (2x-1) en ambos lados:
$\log _{10}\left(2x+1\right)=2\log _{10}\left(3\right)-3\log _{10}\left(2\right)+\log _{10}\left(2x-1\right)$
Aplicamos la propiedad de logaritmos:
$\log _{10}\left(2x+1\right)=\log _{10}\left(9\right)-3\log _{10}\left(2\right)+\log _{10}\left(2x-1\right)$
Aplicamos la propieda de logaritmos:
$\log _{10}\left(2x+1\right)=\log _{10}\left(9\right)-\log _{10}\left(8\right)+\log _{10}\left(2x-1\right)$
Se suma log₁₀ (8) a ambos lados y se aplica propiedad de logaritmos:
$\log _{10}\left(\left(2x+1\right)\cdot \:8\right)=\log _{10}\left(9\left(2x-1\right)\right)$
Como LOg tienen las misma base:
$\left(2x+1\right)\cdot \:8=9\left(2x-1\right)$
Se resuelve:
$16x+8=18x-9$
Nos queda entonces que:
$x=\frac{17}{2}$