Логарифмы. Задание ** фото.

Question 1

Логарифмы. Задание на фото.

Question 2

Решите задачу:

$\frac{\log_336}{\log_43} - \frac{\log_312}{\log_{12}3}= \log_3(3\cdot3\cdot4)\log_34-\log_312\log_312= \\\ =(\log_33+\log_33+\log_34)\log_34-(\log_3(4\cdot3))^2= \\\ =(1+1+\log_34)\log_34-(\log_34+\log_33)^2= \\\ =(2+\log_34)\log_34-(\log_34+1)^2= \\\ =2\log_34+(\log_34)^2-(\log_34)^2-2\log_34-1=-1$

$log_{ \frac{1}{2} }(\cos \frac{ \pi }{6} -\sin \frac{ \pi }{6} )+log_{ \frac{1}{2} }(\cos \frac{ \pi }{6}+\sin \frac{ \pi }{6} )= \\\ =log_{ \frac{1}{2} }(\cos \frac{ \pi }{6} -\sin \frac{ \pi }{6} )(\cos \frac{ \pi }{6}+\sin \frac{ \pi }{6} )= \\\ =log_{ \frac{1}{2} }(\cos^2 \frac{ \pi }{6} -\sin ^2\frac{ \pi }{6} )= log_{ \frac{1}{2} }(\cos \frac{ \pi }{3} )=log_{ \frac{1}{2} }\frac{1}{2}=1 \\\ \log_{13}x=1 \\\ x=13^1=13$

$f(x)=x^{5+\log_2x}-15^{\log_{15}2^{-4}}= x^5\cdot x^{\log_2x}-2^{-4}= \\\ =x^5\cdot x^{\log_2x}-2^{-4}= x^5\cdot x^{\log_2x}-2^{-4} \\\ f( \frac{1}{2} )=(\frac{1}{2})^5\cdot (\frac{1}{2})^{\log_2\frac{1}{2}}-2^{-4}= \\\ =2^{-5}\cdot (\frac{1}{2})^{-1}-2^{-4}=2^{-5}\cdot 2-2^{-4}=2^{-4}-2^{-4}=0$

$11^{\log_{\sqrt{11}}2}+\log_3 \frac{5+2\sqrt{6}}{9}-\log_\frac{1}{\sqrt{3}}( \sqrt{3}-\sqrt{2})= \\\ =\sqrt{11}^{2\log_{ \sqrt{11}}2}+\log_3 \frac{5+2 \sqrt{6}}{9}-\log_{\sqrt{3}}(\frac{1}{\sqrt{3} - \sqrt{2}})= \\\ =\sqrt{11}^{\log_{ \sqrt{11}}2^2}+\log_3 \frac{5+2 \sqrt{6}}{9}-\log_{\sqrt{3}}(\frac{\sqrt{3}+ \sqrt{2}}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})})=$
$2^2+\log_3 \frac{5+2 \sqrt{6}}{9}-\log_{\sqrt{3}}(\sqrt{3}+ \sqrt{2})= \\\ =4+\log_3 \frac{5+2 \sqrt{6}}{9}-\log_3(\sqrt{3}+ \sqrt{2})^2= \\\ =4+\log_3 \frac{5+2 \sqrt{6}}{9}-\log_3(3+2+2\sqrt{6}})= \\\ =4+\log_3 \frac{5+2 \sqrt{6}}{9(5+2\sqrt{6})}=4+\log_3 \frac{1}{9}=4-2=2$

$(m^{ \frac{\log_4n}{\log_2n}}\cdot n^{ \frac{\log_4n}{\log_2n}})^{2\log_{mn}3}= ((mn)^{ \frac{\log_4n}{\log_2n}}}})^{2\log_{mn}3}= \\\ =((mn)^{ \frac{\log_n2}{\log_n4}}}})^{2\log_{mn}3}=((mn)^{\log_42}}})^{2\log_{mn}3}= \\\ =((mn)^{0.5}}}})^{2\log_{mn}3}=(mn}}})^{\log_{mn}3}=3$

$\log_{ \pi ^2} \sqrt{a} =1 \\\ \sqrt{a} = \pi ^2 \\\ a= \pi ^4 \\\ \log_{ \pi ^2} b =1 \\\ b= \pi ^2 \\\ \log_{ \pi ^3}( \frac{ab^3}{ \pi } )=\log_{ \pi ^3}( \frac{ \pi ^4( \pi ^2)^3}{ \pi } )=\log_{ \pi ^3} \pi ^9=3$

$3^{\log_9(x+2 \sqrt{x-2}-1)}+7^{\log_{49}(x-2 \sqrt{x-2}-1)}= \\ =3^{\log_9(x-2+2\sqrt{x-2}+1) }+7^{\log_{49}(x-2-2\sqrt{x-2}+1)}= \\ =3^{\log_9((\sqrt{x-2})^2+2 \sqrt{x-2}+1)}+7^{\log_{49}(( \sqrt{x-2})^2 -2 \sqrt{x-2}+1 )}= \\ =3^{\log_{3^2}( \sqrt{x-2}+1)^2 }+7^{\log_{7^2}(\sqrt{x-2}-1)^2}= \\ =3^{\log_{3}| \sqrt{x-2}+1|}+7^{\log_{7}|\sqrt{x-2}-1| }= \\\ = |\sqrt{x-2}+1|+|\sqrt{x-2}-1|= |\sqrt{2.01-2}+1|+|\sqrt{2.01-2}-1|= \\ =|0.1+1|+|0.1-1|=|1.1|+|-0.9|=1.1+0.9=2$