Question

va rog dau corana urgent​

Answer 1

 

$\displaystyle\bf\\5)\\a)\\lg(x+1)=1\\x+1=10^1\\x+1=10\\x=10-1\\\boxed{\bf~x=9}\\\\b)\\lg(x+1)-2lg(x-1)=1\\lg(x+1)-lg(x-1)^2=1\\\\lg\frac{(x+1)}{(x-1)^2}=1\\\\\frac{(x+1)}{(x-1)^2}=10^1\\\\x+1=10(x-1)^2\\x+1=10(x^2-2x+1)\\10(x^2-2x+1)-(x+1)=0\\10x^2-20x+10-x-1=0\\10x^2-21x+9=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x_{12}=\frac{21\pm\sqrt{441-360}}{20}\\\\x_{12}=\frac{21\pm\sqrt{81}}{20}$

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$\displaystyle\bf\\x_{12}=\frac{21\pm9}{20}\\\\x_1=\frac{21-9}{20}\\\\x_1=\frac{12}{20}\\\\\boxed{\bf~x_1=\frac{3}{5}}\\\\x_2=\frac{21+9}{20}\\\\x_2=\frac{30}{20}\\\\\boxed{\bf~x_2=\frac{3}{2}}$

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$\displaystyle\bf\\c)\\log_3(x-1)=0\\x-1=3^0\\x-1=1\\x=1+1\\\boxed{\bf~x=2}\\\\d)\\ln(x-2)+1=ln\,x\\ln(x-2)-ln\,x=-1\\\\ln\frac{x-2}{x}=-1\\\\\frac{x-2}{x}=e^{-1}\\\\x-2=e^{-1}x\\x-e^{-1}x=2\\x(1-e^{-1})=2\\\\x=\frac{2}{1-e^{-1}}\\\\\\x=\frac{2}{1-\dfrac{1}{e}}\\\\\\x=\frac{2}{\dfrac{e-1}{e}}\\\\\\\boxed{\bf~x=\frac{2e}{e-1}}\\\\\\e)\\log_x(x^2-2x+4)=2\\x^2-2x+4=x^2\\x^2-2x+4-x^2=0\\-2x+4=0\\-2x=-4\\\\x=\frac{-4}{-2}\\\\\boxed{\bf~x=2}$

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$\displaystyle\bf\\f)\\log_2(x+3)-2log_2(x-3)=1\\log_2(x+3)-log_2(x-3)^2=1\\\\log_2\frac{x+3}{(x-3)^2}=1\\\\\frac{x+3}{(x-3)^2}=2^1\\\\x+3=2(x-3)^2\\x+3=2(x^2-6x+9)\\2x^2-13x+15=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x_{12}=\frac{13\pm\sqrt{169-120}}{4}\\\\x_{12}=\frac{13\pm\sqrt{49}}{4}\\\\x_{12}=\frac{13\pm7}{4}\\\\x_1=\frac{13-7}{4}\\\\x_1=\frac{6}{4}\\\\\boxed{\bf~x_1=\frac{3}{2}}\\\\x_=\frac{13+7}{4}\\\\x_2=\frac{20}{4}\\\\\boxed{\bf~x_2=5}$

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$\displaystyle\bf\\g)\\log_2x^2+log_4x=1\\log_2x^2+log_{2^2}x=1\\\\2log_2x+\frac{1}{2} log_2x=1\\\\\left(2+\frac{1}{2}\right)log_2x=1\\\\2,\!5log_2x=1\\\\log_2x=\frac{1}{2,5}\\\\log_2x=\frac{2}{5}\\\\\boxed{\bf~x=2^{^\dfrac{\b2}{\b5}}}\\\\\boxed{\bf~x=\sqrt[\b5]{\bf~x^2}}$

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$\displaystyle\bf\\h)\\log_3x+log_9x+log_{27}x=11\\log_3x+log_{3^2}x+log_{3^3}x=11\\log_3x+\frac{1}{2}log_{3}x+\frac{1}{3}log_{3}x=11\\\\Dam~factor~comun~pe~~log_3x\\\\\left(1+\frac{1}{2}+\frac{1}{3}\right)log_3x=11\\\\\frac{6+3+2}{6}log_3x=11\\\\\frac{11}{6}log_3x=11\\\\log_3x=11:\frac{11}{6}\\\\log_3x=11\times\frac{6}{11}\\\\log_3x=6\\\\x=3^6\\\\\boxed{\bf~x=729}$

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$\displaystyle\bf\\ i)\\lg^2x-3lg\,x+2=0\\Facem~substitutia:~~\boxed{\bf~lg\,x=y}\\\\y^2-3y+2=0\\\\y^2-2y-y+2=0\\\\y(y-2)-1(y-2)=0\\\\(y-2)(y-1)=0\\y_1=1\implies~lg\,x=1 \implies x_1=10^1 \implies \boxed{\bf~x_1=10}\\\\y_2=2\implies~lg\,x=2 \implies x_2=10^2 \implies \boxed{\bf~x_2=100}$

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$\displaystyle\bf\\j)\\log_3(log_4x)=2\\\\log_4x=3^2\\\\log_4x=9\\\\\boxed{\bf~x=4^9}\\\\\boxed{\bf~x=262144}$