Question

Alguém por favor me ajuda nessa questão de matemática?

Answer 1

Olá!

$a) log_{2}( \frac{1}{4} ) \\ {2}^{x} = \frac{1}{4} \\ {2}^{x} = \frac{1}{ {2}^{2} } \\ {2}^{x} = {2}^{ - 2} \\ x = - 2$





$b) log_{3}( \sqrt{3} ) \\ {3}^{x} = \sqrt{3} \\ {3}^{x} = {3}^{ \frac{1}{2} } \\ x = \frac{1}{2}$





$c) log_{8}(16) \\ {8}^{x} = 16 \\ {(2}^{3})^{x} = {2}^{4} \\ {2}^{3x} = {2}^{4} \\ 3x = 4 \\ x = \frac{4}{3}$






$d) log_{4}(128) \\ {4}^{x} = 128 \\ ({2}^{2})^{x} = {2}^{7} \\ {2}^{2x} = {2}^{7} \\ 2x = 7 \\ x = \frac{7}{2}$




$e) log_{36}(6) \\ {36}^{x} = 6 \\ ({6}^{2})^{x} = 6 \\ {6}^{2x} = 6 \\ 2x = 1 \\ x = \frac{1}{6}$




$f) log_{10}(0.01) \\ {10}^{x} = 0.01 \\ {10}^{x} = {10}^{ - 2} \\ x = - 2$




$g) log_{9}( \frac{1}{27} ) \\ {9}^{x} = \frac{1}{27} \\ {( { 3}^{2}) }^{x} = \frac{1}{ {3}^{3} } \\ {3}^{2x} = {3}^{ - 3} \\ 2x = - 3 \\ x = - \frac{ 3}{2}$




$h) log_{0.2}( \sqrt[3]{25} ) \\ {0.2}^{x} = \sqrt[3]{25} \\ { \frac{1}{5} }^{x} = \sqrt[3]{ {5}^{2} } \\ { \frac{1}{5} }^{x} = {5}^{ \frac{2}{3} } \\ { \frac{1}{5} }^{x} = { \frac{1}{5} }^{ - \frac{ 2}{3} } \\ x = - \frac{ 2}{ 3}$





$log_{1.25}(0.64) \\ {1.25}^{x} = 0.64 \\ { \frac{5}{4} }^{x} = \frac{16}{25} \\ { \frac{5}{4} }^{x} = { (\frac{4}{5}) }^{2} \\ { \frac{5}{4} }^{x} = { \frac{5}{4} }^{ - 2} \\ x = - 2$





$j) log_{ \frac{5}{3} }(0.6) \\ { \frac{5}{3} }^{x} = 0.6 \\ { \frac{5}{3} }^{x} = \frac{6}{10} \\ { \frac{5}{3} }^{x} = \frac{3}{5} \\ { \frac{5}{3} }^{x} = { \frac{5}{3} }^{ - 1} \\ x = - 1$





Espero ter ajudado. Bons estudos!!