Question

me ayudan? $log_{2} \sqrt[3]{2^{4}. 16 }$

Answer 1

♛ HØlα!! ✌

Recordemos algunas propiedades de exponentes y logaritmos

             ✅ $\boxed{\boldsymbol{\sqrt[m]{a^n}=a^{\frac{n}{m}}}}$

             ✅ $\boxed{\boldsymbol{\log_n(a\times b)=\log_n{a}+\log_n{b}}}$

             ✅ $\boxed{\boldsymbol{\log_n{a^m}=m(\log_n{a})}}$

             ✅ $\boxed{\boldsymbol{\log_nn=1}}$

Entonces en el problema

                                 $\log_{2}{\sqrt[3]{2^4\cdot16}}=\log_{2}{(2^4\cdot16)^{\frac{1}{3}}}\\\\\\log_{2}{\sqrt[3]{2^4\cdot16}}=\dfrac{1}{3}[\log_{2}{(2^4\cdot16)}]\\\\\\\log_{2}{\sqrt[3]{2^4\cdot16}}=\dfrac{1}{3}[\log_{2}{(2^4)}+\log_{2}{16)}]\\\\\\\log_{2}{\sqrt[3]{2^4\cdot16}}=\dfrac{1}{3}[4\log_{2}{2}+\log_{2}{(2^4)}]\\\\\\\log_{2}{\sqrt[3]{2^4\cdot16}}=\dfrac{1}{3}[4\log_{2}{2}+4\log_{2}{2)}]\\\\\\\log_{2}{\sqrt[3]{2^4\cdot16}}=\dfrac{1}{3}[4(1)+4(1)}]$

                                 $\log_{2}{\sqrt[3]{2^4\cdot16}}=\dfrac{1}{3}(8)\\\\\\\boxed{\boxed{\boldsymbol{\log_{2}{\sqrt[3]{2^4\cdot16}}=\dfrac{8}{3}}}}$