Question

Log(10-x^2)÷log(5-2x)=2

Answer 1

Respuesta:

$x=1$

Explicación paso a paso:

$\frac{\log _{10}\left(10-x^2\right)}{\log _{10}\left(5-2x\right)}=2\\\\\mathrm{Multiplicar\:ambos\:lados\:por\:}\log _{10}\left(5-2x\right)\\\frac{\log _{10}\left(10-x^2\right)}{\log _{10}\left(5-2x\right)}\log _{10}\left(5-2x\right)=2\log _{10}\left(5-2x\right)\\\\\mathrm{Simplificar}\\\log _{10}\left(10-x^2\right)=2\log _{10}\left(5-2x\right)\\\\\mathrm{Aplicar\:las\:propiedades\:de\:los\:logaritmos}:\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)$

$2\log _{10}\left(5-2x\right)=\log _{10}\left(\left(5-2x\right)^2\right)\\\log _{10}\left(10-x^2\right)=\log _{10}\left(\left(5-2x\right)^2\right)\\\\10-x^2=\left(5-2x\right)^2\\\\x=3,\:x=1\\\\\mathrm{La\:solucion\:es}\\\\x=1$

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