Question

¿Me podeis resolver estas ecuaciones?

Answer 1

$\log _{10}\left(x\right)+\log _{10}\left(x-1\right)=\log _{10}\left(12\right)\\\mathrm{Aplicar\:las\:propiedades\:de\:los\:logaritmos}:\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)\\\\\log _{10}\left(x\right)+\log _{10}\left(x-1\right)=\log _{10}\left(x\left(x-1\right)\right)\\\\\log _{10}\left(x\left(x-1\right)\right)=\log _{10}\left(12\right)\\\\x\left(x-1\right)=12\\\\x=4$

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$\log _{10}\left(x+1\right)-\log _{10}\left(x-1\right)=1-\log _{10}\left(6\right)\\\\\log _{10}\left(x+1\right)-\log _{10}\left(x-1\right)+\log _{10}\left(x-1\right)=1-\log _{10}\left(6\right)+\log _{10}\left(x-1\right)\\\\\log _{10}\left(x+1\right)=1-\log _{10}\left(6\right)+\log _{10}\left(x-1\right)\\\\\log _{10}\left(x+1\right)+\log _{10}\left(6\right)=1-\log _{10}\left(6\right)+\log _{10}\left(x-1\right)+\log _{10}\left(6\right)\\\\x=4$

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$4\log _{10}\left(x\right)=2\log _{10}\left(x\right)+\frac{\log _{10}\left(8\right)}{3}\\\\4\log _{10}\left(x\right)-2\log _{10}\left(x\right)=2\log _{10}\left(x\right)+\frac{\log _{10}\left(8\right)}{3}-2\log _{10}\left(x\right)\\\\2\log _{10}\left(x\right)=\frac{\log _{10}\left(8\right)}{3}\\\\\log _{10}\left(x\right)=\frac{\log _{10}\left(2\right)}{2}\\\\x=10^{\frac{\log _{10}\left(2\right)}{2}}$\\\\\x=1.4142

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$\log _{10}\left(3x^2+5x+30\right)-\log _{10}\left(3x+8\right)=1\\\\\log _{10}\left(3x^2+5x+30\right)=1+\log _{10}\left(3x+8\right)\\\\\log _{10}\left(3x^2+5x+30\right)=\log _{10}\left(10\right)+\log _{10}\left(3x+8\right)\\\\\log _{10}\left(3x^2+5x+30\right)=\log _{10}\left(10\left(3x+8\right)\right)\\\\\log _{10}\left(3x^2+5x+30\right)=\log _{10}\left(10\left(3x+8\right)\right)\\\\x=10,\:x=-\frac{5}{3}$

Espero haberte ayudado, me ayudarías marcándo esta respuesta como "la mejor respuesta" ;v