Question

Exercitiile 32 si 33 ...Va rog frumos cei care știți !

Answer 1

32)

$a) \quad \sqrt 3 < 2= \log_{3}9<\log_{3}10 \,\,\Leftrightarrow\,\,\sqrt 3 <2<\log_{3}10 \\ \\ \Rightarrow 2 \in (\sqrt 3, \log_{3}10)\\ \\ b)\quad \sqrt 7 <\sqrt[4]{50}\Big|\verb|^4| \,\,\Leftrightarrow\,\, 49 < 50 \quad (A) \\ \\\log_{2}5 <\sqrt 7\quad ?\\ \\ \dfrac{7}{3}<\sqrt 7 \,\,\Leftrightarrow\,\, \dfrac{49}{9}<7\quad (A) \\ \\\log_{2}5<\dfrac{7}{3} \,\,\Leftrightarrow\,\, 5 <2^{\frac{7}{3}}\,\,\Leftrightarrow\,\, 5^3 < 2^7 \,\,\Leftrightarrow\,\, 125<128\quad (A)$

$\Rightarrow \log_{2}5<\dfrac{7}{3}<\sqrt 7 <\sqrt[4]{50} \Rightarrow \sqrt 7 \in (\log_{2}5 ,\sqrt[4]{50})\\ \\ \\c) \quad \log_{3}2 <\log_{3}3=1<\sqrt 2 \,\,\Leftrightarrow\,\, \log_{3}2 < \sqrt 2 \\ \\ \Rightarrow (-\infty, \log_{3}2) \cap (\sqrt 2,+\infty) = \varnothing$

33)

$a)\quad 4^{\lg 5} = {(10^{\lg 4})}^{\lg 5} = {(10^{\lg 5})}^{\lg 4} = 5^{\lg 4}\\ \\b)\quad \sqrt[4]{2^{\lg 3}}=\sqrt[4]{{(10^{\lg 2})}^{\lg 3}} = \sqrt[4]{{(10^{\lg 3})}^{\lg 2}} = \sqrt[4]{3^{\lg 2}} = \\ \\ = {(3^{\lg 2})}^{\frac{1}{4}} = 3^{\frac{1}{4}\lg 2}= {(3^{\frac{1}{2}})}^{\frac{1}{2}\lg 2} = {(\sqrt 3)}^{\lg 2^{\frac{1}{2}}} = (\sqrt{3})^{\lg \sqrt 2}$