Алгебра: Помогите пожалуйста решить логарифмы. с 036-040

036.
$10^{lg7+lg \frac{2}{7} }=10^{lg(7* \frac{2}{7} )}=10^{lg2}=2$

037.
$16^{log_{4}3- 0.25log_{2}3}= \frac{16^{log_{4}3}}{16^{0.25log_{2}3}}= \frac{(4^2)^{log_{4}3}}{(2^4)^{ \frac{1}{4}log_{2}3 }}= \frac{4^{log_{4}3^2}}{2^{log_{2}3}}= \frac{3^2}{3}=3$

038.
$\frac{1}{3}(1+9^{log_{3}7})^{log_{50}3}= \frac{1}{3}(1+3^{2log_{3}7})^{log_{50}3}= \\ \\ = \frac{1}{3}(1+3^{log_{3}7^2})^{log_{50}3}= \frac{1}{3}(1+7^2)^{log_{50}3}= \\ \\ = \frac{1}{3}(1+49)^{log_{50}3}= \frac{1}{3}*50^{log_{50}3}= \frac{1}{3}*3=1$

039.
$10^{2-lg2}-25^{log_{5}7}= \frac{10^2}{10^{lg2}}-5^{2log_{5}7}= \frac{100}{2}-5^{log_{5}7^2}= \\ =50-7^2=50-49=1$

040.
$2^{2-log_{2}5}+( \frac{1}{2} )^{log_{2}5}= \frac{2^2}{2^{log_{2}5}}+(2^{-1})^{log_{2}5}= \\ \\ = \frac{4}{5}+2^{log_{2}5^{-1}}=0.8+5^{-1}=0.8+ \frac{1}{5}=0.8+0.2=1$