Question

$log(6x - 4) + log(x - 7) = 2$

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Answer 1

Respuesta:

$x=9$

Explicación paso a paso:

$\log _{10}\left(6x-4\right)+\log _{10}\left(x-7\right)=2\\\\\mathrm{Aplicar\:las\:propiedades\:de\:los\:logaritmos}:\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)\\\\\log _{10}\left(\left(6x-4\right)\left(x-7\right)\right)=2\\\\\mathrm{Utilizar\:la\:siguiente\:propiedad\:de\:los\:logaritmos}:\quad \\\mathrm{If}\:\log _a\left(b\right)=c\:\mathrm{then}\:b=a^c\\\\\left(6x-4\right)\left(x-7\right)=10^2\\\\\left(6x-4\right)\left(x-7\right)=100\\\\6x^2-46x+28=100$

$6x^2-46x-72=0\\\\Formula\:general\:para\:ecuaciones\:de\:segundo\:grado:\\\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

1° Valor de x:

$\frac{-\left(-46\right)+\sqrt{\left(-46\right)^2-4\cdot \:6\left(-72\right)}}{2\cdot \:6}\\\\=\frac{46+\sqrt{\left(-46\right)^2+4\cdot \:6\cdot \:72}}{2\cdot \:6}\\\\=\frac{46+\sqrt{3844}}{12}\\\\=\frac{46+62}{12}\\\\=\frac{108}{12}\\\\=9$

2° Valor de x:

$\frac{-\left(-46\right)-\sqrt{\left(-46\right)^2-4\cdot \:6\left(-72\right)}}{2\cdot \:6}\\\\=\frac{46-\sqrt{\left(-46\right)^2+4\cdot \:6\cdot \:72}}{2\cdot \:6}\\\\=\frac{46-\sqrt{3844}}{12}\\\\=\frac{46-62}{12}\\\\=-\frac{16}{12}\\\\=-\frac{4}{3}$

$\mathrm{Verificar\:las\:soluciones\:sustituyendolas\:en\:}\log _{10}\left(6x-4\right)+\log _{10}\left(x-7\right)=2\\\\\mathrm{Sustituir}\:x=9\\\log _{10}\left(6\cdot \:9-4\right)+\log _{10}\left(9-7\right)=2\\\\2=2\\\\\mathrm{Verdadero}\\\\\\\mathrm{Sustituir}\:x=-\frac{4}{3}\\\\\log _{10}\left(6\left(-\frac{4}{3}\right)-4\right)+\log _{10}\left(\left(-\frac{4}{3}\right)-7\right)=2\\\\Falso\\\\\mathrm{La\:solucion\:es}\\\\x=9$

Answer 2

Respuesta:

x = 9

Explicación paso a paso:

Aplicar propiedad:

$log(x) \: + \: log(y) = log(x \cdot{y})$

Entonces:

$log(6x - 4) \: + \: log(x - 7) = 2$

$log(6x - 4)(x - 7) = 2$

Aplicar propiedad:

$log(a) = log_{10}(a)$

$log_{b}a = c \: \: \: \: \: \: \to \: \: \: \: \: \: \: {b}^{c} = a$

Entonces:

$log(6x - 4)(x - 7) = 2$

$log _{10}(6x - 4)(x - 7) = 2$

${10}^{2} = (6x - 4)(x - 7)$

$100 = {6x}^{2} - 42x - 4x + 28$

$100 = {6x}^{2} - 46x + 28$

$\frac{100}{2} = \frac{{6x}^{2}}{2} - \frac{46x}{2} + \frac{28}{2}$

$50 = {3x}^{2} - 23x + 14$

$0 = {3x}^{2} - 23x + 14 - 50$

$0 = {3x}^{2} - 23x - 36$

$0 = (3x + 4)(x - 9)$

$(3x + 4) = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (x - 9) = 0$

$3x = - 4 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = 9$

$x_{1} = - \frac{4}{3} \: \: \: \: \: \: \: \: \: \: \: \: \: \: x_{2}= 9$

"X" no puede ser negativo, entonces la única solución es 9