Question

log(28 - x^3)- 3log (4 - x)=0 como se resuelve?

Answer 1

SOLUCIÓN

Hola!!   (⌐■_■)

Utilizaremos las siguientes propiedades de logaritmos

                                     $1. \boxed{\boldsymbol{\log_{b}N = x \Rightarrow b^{x} = N }}\\\\\\2. \boxed{\boldsymbol{\log_{b}X - \log_{b}Y = \log_{b}(\frac{X}{Y}) }}\\\\\\3. \boxed{\boldsymbol{\log_{b}N^{x} = x \log_{b}N}}$

Entonces

$\log(28 - x^{3}) - 3 \log(4-x) = 0 \\\\\log(28 - x^{3}) - \log(4-x)^{3} = 0\\\\\log ( \frac{28 - x^{3}}{(4-x)^{3}} ) = 0 \\\\\frac{28 - x^{3}}{(4-x)^{3}}= 10^{0} \\\\28 - x^{3} = (4-x)^{3}\\\\28 - \cancel{x^{3}} = 4^{3} - \cancel{x^{3}} - 3(4)(x)(4-x)\\\\28 = 64 - 12x(4-x)\\\\12x(4-x) = 36\\\\x(4-x) = 3 \\\\4x - x^{2} = 3\\\\x^2 - 4x + 3 = 0\\\\(x-3)(x-1) = 0\\\\\boxed{x = 3} \: \: \: o \: \: \: \boxed{x = 1}$