Question

Hallar el valor de “x”: 2log3 27=4log3 9-3x

Answer 1

2log3 27=4log3 9-3x:

$\mathrm{Intercambiar\:lados}\\\\4\log _{10}\left(39\right)-3x=2\log _{10}\left(327\right)\\\\\mathrm{Restar\:}4\log _{10}\left(39\right)\mathrm{\:de\:ambos\:lados}\\\\4\log _{10}\left(39\right)-3x-4\log _{10}\left(39\right)=2\log _{10}\left(327\right)-4\log _{10}\left(39\right)\\\\Simplificar\\\\4\log _{10}\left(39\right)-3x-4\log _{10}\left(39\right)=2\log _{10}\left(327\right)-4\log _{10}\left(39\right)$$\mathrm{Simplificar\:}4\log _{10}\left(39\right)-3x-4\log _{10}\left(39\right):\quad -3x\\\\\mathrm{Simplificar\:}2\log _{10}\left(327\right)-4\log _{10}\left(39\right):\quad \log _{10}\left(\frac{11881}{257049}\right)\\\\-3x=\log _{10}\left(\frac{11881}{257049}\right)\\\\-3x=\log _{10}\left(\frac{11881}{257049}\right)\\\\\mathrm{Dividir\:ambos\:lados\:entre\:}-3\\\\\frac{-3x}{-3}=\frac{\log _{10}\left(\frac{11881}{257049}\right)}{-3}\\\\\mathrm{Simplificar}$

$x=-\frac{\log _{10}\left(\frac{11881}{257049}\right)}{3}$