Question

Пошагово решить логарифмы:

Answer 1

Объяснение:

a)

$\frac{2*log_{7}6-log_{7} 3 }{log_{7}144 } =\frac{log_{7}6^{2}-log_{7}3 }{log_{7}12^{2} } =\frac{log_{7}36-log_{7}3 }{2*log_{7}12 } =\frac{log_{7}\frac{36}{3} }{2*log_{7}12 } =\frac{log_{7}12 }{2*log_{7}12 } =\frac{1}{2} .$

б)

$\frac{log_{2}(2-\sqrt{3}) }{log_{4} (2+\sqrt{3}) }=\frac{log_{2}(2-\sqrt{3})+log_{2} (2+\sqrt{3}) -log_{2}(2+\sqrt{3} ) }{\frac{1}{2} *log_{2} (2+\sqrt{3}) }=\frac{2*log_{2}((2-\sqrt{3})(2+\sqrt{3}))-log_{2}(2 +\sqrt{3}) }{log_{2}(2+\sqrt{3}) } =\\ =\frac{2*log_{2} (2^{2}-(\sqrt{3})^{2})-log_{2}(2+\sqrt{3}) }{log_{2} (2+\sqrt{3}) } =\frac{2*log_{2} (4-3)-log_{2}(2+\sqrt{3}) }{log_{2} (2+\sqrt{3})} =\frac{2*log_{2} 1-log_{2}(2+\sqrt{3}) }{log_{2} (2+\sqrt{3}) } =$

$=\frac{2*0-log_{2}(2+\sqrt{3}) }{log_{2} (2+\sqrt{3}) } =-1.$

в)

$log_{\frac{1}{4} } (log_{2} 3*log_{3}2^{2} )^{2} =log_{\frac{1}{4} } (log_{2} 3*log_{3}4)^{2} =log_{\frac{1}{4} } (2*log_{2} 3*log_{3}2)^{2} =\\=2* log_{\frac{1 }{4} } (2*log_{2} 3*log_{3}2)=2*log_{2^{-2} } (2*\frac{log_{2} 3}{log_{2} 3}) =-\frac{2} {2}log_{2}(2*1) =-log_{2}2=-1.$