Question

ajutoooooor, dau coroana ​

Answer 1

Explicație pas cu pas:

$log_{2}(3) = \dfrac{ ln(3) }{ ln(2) } = a \iff ln(3) = a \cdot ln(2)$

$log_{24}(54) = \dfrac{ ln(2 \cdot {3}^{3} ) }{ ln( {2}^{3} \cdot 3) } = \dfrac{ ln(2) + ln({3}^{3}) }{ ln( {2}^{3}) + ln(3) } \\ = \dfrac{ ln(2) + 3ln(3) }{ 3ln(2) + ln(3) } = \dfrac{ ln(2) + 3a \cdot ln(2) }{ 3ln(2) + a \cdot ln(2) } \\ = \dfrac{ ln(2) \cdot (1 + 3a)}{ ln(2) \cdot (3 + a)} = \dfrac{1 + 3a}{3 + a}$

$\implies log_{24}(54) = \dfrac{1 + 3a}{3 + a}$