Question

logx na base 2 + log x na base 4 + log x na base 8

Answer 1

$\mathrm{\log_2{x}+\log_4{x}+\log_8{x}}\\\\ \textrm{Pela propriedade de mudan\c{c}a de base, teremos que:}\\\\ \mathrm{\dfrac{\log{x}}{\log{2}}+\dfrac{\log{x}}{\log{4}}+\dfrac{\log{x}}{\log{8}}=\dfrac{\log{x}}{\log{2}}+\dfrac{\log{x}}{\log{2^2}}+\dfrac{\log{x}}{\log{2^3}}}\\\\ \textrm{Pela propriedade do log da pot\^encia, teremos que:}\\\\ \mathrm{\dfrac{\log{x}}{\log{2}}+\dfrac{\log{x}}{2.\log{2}}+\dfrac{\log{x}}{3.\log{2}}\ \to\ MMC=6.log{2}}$

$\mathrm{\dfrac{6.\log{x}+3.\log{x}+2.\log{x}}{6.\log{2}}=\dfrac{11.\log{x}}{6.\log{2}}}\\\\\\ \textrm{A partir da\'i, temos 3 formas de simplifica\c{c}\~ao:}\\\\ \mathrm{\mathbf{I.}\ \dfrac{11.\log{x}}{6.\log{2}}=\dfrac{\log{x^{11}}}{\log{2^6}}=\log_{64}{x^{11}}}\\\\\\ \mathrm{\mathbf{II.}\ \dfrac{11.\log{x}}{6.\log{2}}=\dfrac{\log{x^{\frac{11}{6}}}}{\log{2}}=\log_2{x^{\frac{11}{6}}}}\\\\\\ \mathrm{\mathbf{III.}\ \dfrac{11.\log{x}}{6.\log{2}}=\dfrac{11.\log_2{x}}{6}}$