Question

Me ajudem por favor!

Answer 1

1 -

$log_{125}625 ===\ \textgreater \ 125^x = 625 \\ (5^3)^x = 5^4 \\ \\ 3x = 4 \\ \\ x = \frac{4}{3} \\ \\ log \sqrt[4]{1000} ==\ \textgreater \ log10^{ \frac{3}{4}} ==\ \textgreater \ \frac{3}{4}.log10 = \frac{3}{4} \\ \\ log_{81} \frac{1}{243} ==\ \textgreater \ log_{3^4} 243^{-1} ==\ \textgreater \ \frac{1}{4}.log_3 (3^5)^{-1} \\ \\ \frac{1}{4}.log_3 3^{-5} ==\ \textgreater \ \frac{-5}{4}.log_33 \\ \\ \frac{-5}{4}$


$log_{ \frac{27}{8} } \frac{64}{729} ===\ \textgreater \ (\frac{27}{8})^x = \frac{64}{729} \\ \\ (\frac{3^3}{2^3})^x = \frac{2^6}{3^6} ==\ \textgreater \ (\frac{3}{2})^{3x} = (\frac{3}{2})^-6 \\ \\ 3x= - 6 \\ x = -2$

2 - 

$log_610 = log_6(2.5) = log_62 + log_65 ==\ \textgreater \ 0,386 + 0,898 = 1,284 \\ \\ log_25 ==\ \textgreater \ \frac{log_65}{log_62} = \frac{0,898}{0,386} \approx 2,3 \\ \\ log_62,5 ==\ \textgreater \ log_6 \frac{25}{10} ==\ \textgreater \ log_625 - log_610 \\ \\ log_65^2 - 1,284 ==\ \textgreater \ 2.log_65 - 1,284 \\ 2.0,898 - 1,284 = 0,512 \\ \\ log_620 ==\ \textgreater \ log_6(2^2.5) ==\ \textgreater \ log_62^2 + log_65 \\ \\ 2.log_62 + 0,898 ==\ \textgreater \ 2.0,386 + 0,898 = 1,67$


$log_6\sqrt{5} ==\ \textgreater \ log_65^{ \frac{1}{2}} ==\ \textgreater \ \frac{1}{2}.log_65 \\ \\ \frac{1}{2}.0,898 = 0,449$