Question

Sa se logaritmeze expresiile:​

Answer 1

Explicație pas cu pas:

a)

$ln(E) = ln( {41}^{2} \sqrt[13]{41 \cdot {37}^{5} } ) = ln( {41}^{2} \cdot {41}^{ \frac{1}{13} } \cdot {37}^{ \frac{5}{13} }) =$

$= ln({41}^{ \frac{27}{13} } \cdot {37}^{ \frac{5}{13} }) = ln({41}^{ \frac{27}{13} }) + ln({37}^{ \frac{5}{13} })$

$= \dfrac{27}{13} ln \ 41 + \dfrac{5}{13} ln \ 37$

b)

$ln(E) = ln\dfrac{ {31}^{3} \cdot \sqrt[7]{41 \cdot {33}^{4} } }{ {17}^{2} \cdot \sqrt[3]{ {23}^{2} \cdot 29} } =$

$= ln({31}^{3} \cdot {41}^{\frac{1}{7}} \cdot {33}^{\frac{4}{7}}) - ln({17}^{2} \cdot {23}^{ \frac{2}{3} } \cdot {29}^{ \frac{1}{3} } )$

$= (ln{31}^{3} + ln {41}^{\frac{1}{7}} + ln {33}^{\frac{4}{7}}) - (ln{17}^{2} + ln {23}^{ \frac{2}{3} } + ln {29}^{ \frac{1}{3} } )$

$= 3ln \ 31 + \dfrac{1}{7}ln \ 41 + \dfrac{4}{7}ln \ 33 - 2ln \ 17 - \dfrac{2}{3}ln \ 23 - \dfrac{1}{3}ln \ 29$

Answer 2

Răspuns:

Explicație pas cu pas: