Ln(x-2)+ln(2x-3)=2lnx

Ответ:

meganldale15

01.02.2022

$ln(x-2)+ln(2x-3)=2lnx\\\\D:x-2 0\ \wedge\ 2x-3 0\ \wedge\ x 0\\x 2\ \wedge\ x 1.5\ \wedge\ x 0\\x\in(2;\ \infty)$

$ln[(x-2)(2x-3)]=lnx^2\\\\ln(2x^2-3x-4x+6)=lnx^2\\\\ln(2x^2-7x+6)=lnx^2\iff2x^2-7x+6=x^2\\\\2x^2-x^2-7x+6=0\\\\x^2-7x+6=0\\\\x^2-x-6x+6=0\\x(x-1)-6(x-1)=0\\(x-1)(x-6)=0\iff x-1=0\ \vee\ x-6=0\\\\x=1\notin D;\ x=6\in D\\\\Solution:x=6.$

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Ответ:

edeliz5033

01.02.2022

Mathematics

$\ln(x-2)+\ln(2x-3)=2\ln x \\
D:x-20 \wedge 2x-3 0 \wedge x0\\
D:x2 \wedge 2x3 \wedge x0\\
D:x2 \wedge x\frac{3}{2}\\
D:x2\\
\ln(x-2)(2x-3)=\ln x^2\\
2x^2-3x-4x+6=x^2\\
x^2-7x+6=0\\
x^2-x-6x+6=0\\
x(x-1)-6(x-1)=0\\
(x-6)(x-1)=0\\
x=6 \vee x=1\\
1\not \in D\Rightarrow x=6


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Ответ:

bncook

28.01.2020

Mathematics

The answer to your question is option b because opposite angles in a quadrilateral are equal to one another

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