Question

holi, alguien me ayuda¿?
verificar las igualdades logarítimicas:

Answer 1

Respuesta:

↓

Explicación paso a paso:

$\log _3\left(\frac{\frac{1}{3}\sqrt{27}}{\sqrt[3]{3}\sqrt[5]{1^7}}\right)=0\\\\\mathrm{Simplificar\:}\log _3\left(\frac{\frac{1}{3}\sqrt{27}}{\sqrt[3]{3}\sqrt[5]{1^7}}\right):\\\\\log _3\left(\frac{\frac{1}{3}\sqrt{27}}{\sqrt[3]{3}\sqrt[5]{1^7}}\right)\\\\\mathrm{Aplicar\:la\:regla}\:1^a=1\\\\=\log _3\left(\frac{\frac{1}{3}\sqrt{27}}{\sqrt[3]{3}\sqrt[5]{1}}\right)\\\\\mathrm{Aplicar\:la\:regla}\:\sqrt[n]{1}=1\\\\\log _3\left(\frac{\sqrt{27}\frac{1}{3}}{1\cdot \:\sqrt[3]{3}}\right)$

$\log _3\left(\frac{\sqrt{3}}{\sqrt[3]{3}}\right)\\\\\mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}\\\\\log _3\left(\frac{3^{\frac{1}{2}}}{3^{\frac{1}{3}}}\right)\\\\\mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \frac{x^a}{x^b}=x^{a-b}\\\\\log _3\left(3^{\frac{1}{2}-\frac{1}{3}}\right)\\\\\log _3\left(3^{\frac{1}{6}}\right)\\\\\mathrm{Aplicar\:las\:leyes\:de\:los\:exponentes}:\quad \:a^{\frac{1}{n}}=\sqrt[n]{a}$

$\log _3\left(\sqrt[6]{3}\right)$

$\log _3\left(\sqrt[6]{3}\right)=0\\\\\mathrm{Los\:lados\:no\:son\:iguales}\\\\\mathrm{Falso}$

$\log _5\left(\frac{1^5\sqrt{25}}{\sqrt[4]{5}\sqrt[3]{125}}\right)=\frac{1}{4}\\\\\mathrm{Simplificar\:}\log _5\left(\frac{1^5\sqrt{25}}{\sqrt[4]{5}\sqrt[3]{125}}\right):\\\\\log _5\left(\frac{1^5\sqrt{25}}{\sqrt[4]{5}\sqrt[3]{125}}\right)\\\\\mathrm{Aplicar\:la\:regla}\:1^a=1\\\\=\log _5\left(\frac{1\cdot \sqrt{25}}{\sqrt[4]{5}\sqrt[3]{125}}\right)\\\\=\log _5\left(\frac{1}{\sqrt[4]{5}}\right)\\\\=\log _5\left(\frac{5^{\frac{3}{4}}}{5}\right)$

$\log _5\left(\frac{5^{\frac{3}{4}}}{5}\right)=\frac{1}{4}\\\\\mathrm{Los\:lados\:no\:son\:iguales}\\\\Falso$