Question

Am nevoie urgent! Va rog ajutati-mă!

Answer 1

 

$\displaystyle\bf\\1a)\\log_{(x+1)}2\\(x+1)>0~\implies~x>-1\\(x+1)\neq 1~\implies~x\neq 0\\~~~\implies~x\in (-1,~\infty)\backslash \{-1\}\\\\1b)\\log_5(x^2-5x+4)\\\\x^2-5x+4=x^2-1x-4x+4=x(x-1)-4(x-1)=(x-1)(x-4)\\x_1=1~si~x_2=4\\Intre~radacini~are~semn~contrar~lui~a.Conditie:~~~x^2-5x+4>0\\\implies~~x\in(-\infty,~1)\bigcup~(4,~+\infty)$

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$\displaystyle\bf\\2a)\\log_26+log_23-log_29=log_2\Big(\frac{6\times3}{9}\Big)=log_2\Big(\frac{18}{9}\Big)=log_22=1\\\\2b)\\\sqrt{log_39+log_99+log_33}=\sqrt{2+1+1}=\sqrt{4}=2\\\\2c)\\log_5(4-log_28)=log_5(4-3)=log_5(1)=0$

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$\displaystyle\bf\\3a)\\ E=\frac{log_2x+log_2x^3}{log_3x^2-log_3x}\\\\\\E=\frac{log_2(x\cdot x^3)}{log_3\Big(\dfrac{x^2}{x}\Big)}\\\\\\E=\frac{log_2(x^4)}{log_3x}\\\\\\E=4\cdot\frac{log_2 x}{log_3x}~~Rasturnam~fractia\\\\\\E=4\cdot\frac{log_x 3}{log_x2}\\\\\\E=4log_23\\\\\implies~E~nu~depinde~de~x.$

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$\displaystyle\bf\\3b)\\log_2\Big(\frac{1}{2}\Big)+log_2\Big(\frac{2}{3}\Big)+...+log_2\Big(\frac{7}{8}\Big)=\\\\=log_2\Big(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\Big)=\\\\=log_2\Big(\frac{1}{8}\Big)=log_2\Big(\frac{1}{2^3}\Big)=log_2\Big(2^{-3}\Big)=-3 \in Z$

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$\displaystyle\bf\\3c)~~Aratam~ca:\\log_3{(24)}=1+3a,~daca~a=log_32\\\\log_3{(24)}=1+3\cdot log_32\\\\log_3{(24)}=log_33+3\cdot log_32\\\\log_3{(24)}=log_33+log_32^3\\\\log_3{(24)}=log_3{(3\cdot 2^3)}\\\\log_3{(24)}=log_3{(3\cdot 8)}\\\\log_3{(24)}=log_3{(24)}~~~Egalitate~demonstrata.$

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$\displaystyle\bf\\4a)\\lg\,x=lg\,5-2lg\,3\\lg\,x=lg\,5-lg\,3^2\\lg\,x=lg\,5-lg\,9\\lg\,x=lg \frac{5}{9}\\\\x=\frac{5}{9}\\\\\\4b)\\log_3(x)=log_3\Bigg(\sqrt{3-\sqrt{5}} \Bigg)+log_3\Bigg(\sqrt{3+\sqrt{5}} \Bigg)\\\\\\log_3(x)=log_3\left[\Bigg(\sqrt{3-\sqrt{5}} \Bigg)\cdot\Bigg(\sqrt{3+\sqrt{5}}\Bigg)\right]\\\\\\log_3(x)=log_3\sqrt{\Big({3-\sqrt5} \Big)\cdot\Big(3+\sqrt{5}\Big)}\\\\\\log_3(x)=log_3\sqrt{{3^2-\Big(\sqrt5} \Big)^2}\\\\\\log_3(x)=log_3\sqrt{9-5}\\\\x=\sqrt{9-5}=\sqrt{4}=2$