Question

Задание по математике в фотографии пожалуйста решите

Answer 1

Ответ:

$1) log3( {3}^{ - 3} ) = - 3$

$2)1$

$3) log {5}^{ - 1} ( {5}^{3} ) = - 1 \times 3 \times log5(5) = - 3$

$4) { \frac{1}{3} }^{ log \frac{1}{3} ( {7}^{2} ) } = {7}^{2} = 49$

$5)12 \times {12}^{ log12(7) } = 12 \times 7 = 84$

$6) log0.1(5 \times 2) = log0.1(10) = - 1$

$7) log3( \frac{1}{6} \div 40.5) = log3( \frac{1}{243} ) = log3( {3}^{ - 5} ) = - 5$

$8) log2(56) + log2( {12}^{2} ) - log2(63) = log2( \frac{56 \times 144}{63} ) = log_{2}(128) = 7$

2 задание

$1)2x + 3 = {4}^{3} \\ 2x + 3 = 64 \\ 2x = 61 \\ x = \frac{61}{2}$

$2) log_{3}(x) + log_{ {3}^{2} }(x) + log_{ {3}^{3} }(x) = \frac{11}{12 } \\ log_{3}(x) + \frac{1}{2} log_{3}(x) + \frac{1}{3} log_{3}(x) = \frac{11}{12} \\ \frac{11}{6} log_{3}(x) = \frac{11}{12} \\ log_{3}(x) = \frac{11}{12} \times \frac{6}{11} \\ x = {3}^{ \frac{1}{2} }$

$3 ) {5}^{x} = t \\ {t}^{2} + 2t - 15 = 0 \\ t1 = 3 \\ t2 = - 5 \\ {5}^{x} = 3 \\ x1 = log_{5}(3) \\ {5}^{x} = - 5 \\ net \: resheniya$