Question

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Answer 1

$\it a)\ log_2 2\sqrt{32}==log_2 2\cdot32^{\frac{1}{2}}=log_2 2\cdot(2^5)^\frac{1}{2}=log_2 2\cdot2^{\frac{5}{2}}=log_2 2+log_2 2^{\frac{5}{2}}=\\ \\ \\ =1+\dfrac{5}{2}\cdotlog_2 2=1+\dfrac{5}{2}=1+2,5=3,5\\ \\ \\ b)\ log_3(\sqrt[3]{3}\cdot27)=log_3(3^{\frac{1}{3}}\cdot3^3)=log_3 3^{\frac{1}{3}+3}=log_3 3^{\frac{10}{3}}=\dfrac{10}{3}\cdot log_3 3=\dfrac{10}{3}$

$\it c)\ log_{\frac{1}{\sqrt{5}}}\sqrt{125}=\dfrac{log_5\sqrt{125}}{log_5 \frac{1}{\sqrt5}}=\dfrac{log_5\sqrt{5^3}}{log_5(\sqrt5)^{-1}}=\dfrac{log_5{5^\frac{3}{2}}}{log_55^{-\frac{1}{2}}}=\dfrac{\dfrac{3}{2}}{-\dfrac{1}{2}}=-\dfrac{3}{2}\cdot\dfrac{2}{1}=-3$

$\it g)\ log_35+log_35,4=log_3 5\cdot5,4=log_327=log_3 3^3=3log_33=3\\ \\ \\i)\ log_612+log_65-log_610=log_6\dfrac{12\cdot5}{10}=log_66=1$