Question

Ma poate ajuta si pe mine cineva dau puncte

Answer 1

Explicație pas cu pas:

1.$log_{2}(64) = log_{2}( {2}^{6} ) = 6$

$log_{0.1}(10) = log_{ {10}^{ - 1} } ( {10} ) = - 1$

$log_{ \frac{1}{5} }(0.008) = log_{ {5}^{ - 1} }( {5}^{ - 3} ) = 3$

2.

$log_{6}(1) = 0$

$log_{3}( \frac{1}{27} ) = log_{3}({3}^{ - 3} ) = - 3$

$log_{5}(125) = log_{5}( {5}^{3} ) = 3$

$log_{8}(0.025) = log_{ {2}^{3} }( {40}^{ - 1} ) = - \frac{1}{3 } \times log_{2}(40) = - \frac{1}{3} log_{2}(8 \times 5) = - \frac{1}{3} \times ( log_{2}(8) + log_{2}(5) ) = - \frac{1}{3 } \times ( log_{2}( {2}^{3}) + log_{2}(5) ) = - \frac{1}{3} \times (3 + log_{2}(5) = - 1 - \frac{1}{3} \times log_{2}(5)$

3.

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

$log_{10}( \sqrt{10} ) = log_{10}( {10}^{ \frac{1}{2} } ) = \frac{1}{2}$

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

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