Question

va rog ajutati-ma, dau coroana!​

Answer 1

$\displaystyle a)~ log_2x+log_4x+log_8x=5,5 \\\\~Formula~pentru~schimbarea~bazei~unui~logaritm:~\boxed{log_ax=\frac{log_bx}{log_ba} }\\\\ log_4x=\frac{log_2x}{log_24}=\frac{log_2x}{\underbrace{log_22^2} }=\frac{log_2x}{2\underbrace{log_22}} =\frac{log_2x}{2} \\~~~~~~~~~~~~\boxed{log_ax^n=nlog_ax}~\boxed{log_aa=1}\\\\ log_8x=\frac{log_2x}{log_28} =\frac{log_2x}{log_22^3} =\frac{log_2x}{3log_22}=\frac{log_2x}{3}$

$\displaystyle log_2x+log_4x+log_8x=5,5 \Rightarrow log_2x+\frac{log_2x}{2} +\frac{log_2x}{3} =5,5 \Rightarrow \\\\ \Rightarrow 6log_2x+3log_2x+2log_2x=5,5 \cdot6 \Rightarrow 11log_2x=33 \Rightarrow log_2x=\frac{33}{11} \Rightarrow \\\\ \Rightarrow \underbrace{log_2x=3}\Rightarrow x=2^3 \Rightarrow x=8\\\boxed{log_ax=b\Rightarrow x=a^b}\\\\ x\in\{8\}$

$\displaystyle b)~\underbrace{log_7(x-2)+log_7(2x+1)}=1 \Rightarrow log_7(x-2)(2x+1)=1 \Rightarrow \\~~~~~~ \boxed{log_ax+log_ay=log_axy}\\\\ \Rightarrow log_7\Big(2x^2+x-4x-2\Big)=log_77 \Rightarrow 2x^2+x-4x-2=7 \Rightarrow \\\\ \Rightarrow 2x^2-3x-2-7=0 \Rightarrow 2x^2-3x-9=0 \\\\ \Delta=(-3)^2-4 \cdot 2 \cdot (-9)=9+72=81 > 0$

$\displaystyle x_1=\frac{-(-3)-\sqrt{81} }{2\cdot2}=\frac{3-9}{4} =-\frac{6}{4} =-\frac{3}{2} \\\\x_2=\frac{-(-3)+\sqrt{81} }{2\cdot2}=\frac{3+9}{4} =\frac{12}{4} =3 \\\\ x\in\{3\}$

$\displaystyle c)~log_3\frac{x-2}{x-1} -1=log_3\frac{3x-7}{3x-1} \Rightarrow \underbrace{log_3\frac{x-2}{x-1} -log_33}=log_3\frac{3x-7}{3x-1} \Rightarrow \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{log_ax-log_ay=log_a\frac{x}{y} }$

$\displaystyle \Rightarrow log_3\frac{\cfrac{x-2}{x-1} }{3} =log_3\frac{3x-7}{3x-1} \Rightarrow \frac{\cfrac{x-2}{x-1} }{3} =\frac{3x-7}{3x-1} \Rightarrow \frac{x-2}{x-1}\cdot \frac{1}{3} =\frac{3x-7}{3x-1} \Rightarrow \\\\ \Rightarrow \frac{x-2}{3x-3} =\frac{3x-7}{3x-1}\Rightarrow (x-2)(3x-1)=(3x-3)(3x-7) \Rightarrow \\\\ \Rightarrow 3x^2-x-6x+2=9x^2-21x-9x+21 \Rightarrow \\\\ \Rightarrow 3x^2-x-6x+2-9x^2+21x+9x-21=0 \Rightarrow -6x^2+23x-19=0$

$\displaystyle \Delta=23^2-4 \cdot (-6) \cdot (-19)=529-456=73 > 0\\\\ x_1=\frac{-23-\sqrt{73} }{2 \cdot (-6)} =\frac{23+\sqrt{73} }{12} ;~x_2=\frac{-23+\sqrt{73} }{2\cdot (-6)} =\frac{23-\sqrt{73} }{12} \\\\ x\in \Bigg\{\frac{23+\sqrt{73} }{12} \Bigg\}$

$\displaystyle d)~log_5\frac{x-6}{2x+1} +log_5\frac{2x+1}{x+1} =1 \Rightarrow log_5\Bigg(\frac{x-6}{2x+1} \cdot \frac{2x+1}{x+1} \Bigg)=log_55 \Rightarrow \\\\ \Rightarrow log_5\frac{x-6}{x+1} =log_55 \Rightarrow \frac{x-6}{x+1} =5 \Rightarrow x-6=5(x+1) \Rightarrow x-6=5x+5 \Rightarrow \\\\ \Rightarrow x-5x=5+6 \Rightarrow -4x=11 \Rightarrow x=-\frac{11}{4} \\\\ x\in \Bigg\{-\frac{11}{4}\Bigg\}$

$\displaystyle e)~log_5(x+1)+log_5(x+2)+log_5(x+3)=3log_52+log_53 \Rightarrow \\\\ \Rightarrow log_5((x+1)(x+2)(x+3))=log_52^3+log_53\Rightarrow \\\\ \Rightarrow log_5\Big(\Big(x^2+3x+2\Big)(x+3)\Big)=log_58+log_53 \Rightarrow\\\\ \Rightarrow log_5\Big(x^3+6x^2+11x+6\Big)=log_5(8\cdot 3)\Rightarrow\\\\\Rightarrow x^3+6x^2+11x+6=24 \Rightarrow x^3+6x^2+11x+6-24=0 \Rightarrow \\\\ \Rightarrow x^3+6x^2+11x-18=0 \Rightarrow (x-1)\Big(x^2+7x+18\Big)=0$

$\displaystyle x-1=0 \Rightarrow x=0+1\Rightarrow x=1 \\\\ x^2+7x+18=0 \\\\ \Delta=7^2-4 \cdot 1 \cdot 18=49-72=-23 \Rightarrow Nu~exista~solutii~reale.\\\\ x\in\{1\}$