Răspuns:
Explicație pas cu pas:
[tex]\log_{60}{24}=\log_{60}(2^3\cdot 3)=\log_{60}(2^{3})+\log_{60}{3}=3\log_{60}2+\log_{60}3=\\=\dfrac{3}{\log_{2}{60}}+\dfrac{1}{\log_360}=\dfrac{3}{\log_{2}(3\cdot 2\cdot 10)}+\dfrac{1}{\log_3(2\cdot 3\cdot 10)}=\\=\dfrac{3}{\log_23+\log_22+\log_210}+\dfrac{1}{\log_32+\log_33+\log_310}=\\=\dfrac{3}{\frac{\lg 3}{\lg 2}+1+\frac{1}{\lg 2}}+\dfrac{1}{\frac{\lg 2}{\lg 3}+1+\frac{1}{\lg 3}}=\dfrac{3}{\frac{b}{a}+1+\frac{1}{a}}+\dfrac{1}{\frac{a}{b}+1+\frac{1}{b}}=\\=\dfrac{3a}{b+a+1}+\dfrac{b}{a+b+1}=\dfrac{3a+b}{a+b+1}[/tex]
