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konrad509
[tex]D:5x-2>0 \wedge x>0 \wedge x-1>0\\D:5x>2 \wedge x>0 \wedge x>1\\D: x>\frac{2}{5} \wedge x>1\\D:x>1\\\log_2(5x-2)-\log_2x-\log_2(x-1)=2\\\log_2\frac{5x-2}{x(x-1)}=\log_24\\\frac{5x-2}{x(x-1)}=4\\4x(x-1)=5x-2\\4x^2-4x=5x-2\\4x^2-9x+2=0\\4x^2-x-8x+2=0\\x(4x-1)-2(4x-1)=0\\(x-2)(4x-1)=0\\x=2 \vee x=\frac{1}{4}\\\frac{1}{4}\not \in D \Rightarrow \boxed{x=2}[/tex]
superkoopasirf
[tex]\log_2(5x - 2) - \log_2 x - \log_2(x - 1) = 2[/tex]
[tex]\log_2 ( \frac{5x - 2}{x(x-1)} ) = 2[/tex]
[tex]\log_2 ( \frac{5x - 2}{x(x-1)} ) = \log_2 4[/tex]
[tex]\frac{5x - 2}{x(x-1)} = 4[/tex]
[tex]5x - 2 = 4x(x - 1)[/tex]
[tex]5x - 2 = 4x^2 - 4x[/tex]
[tex]0 = 4x^2 - 9x + 2[/tex]
[tex]0 = (4x - 1)(x - 2)[/tex]
[tex]\implies x = 2 \text{ or } x = \frac{1}{4} [/tex]

By putting one quarter into the equation we can see that it cannot work in any way involving real numbers as [tex]\log_2 ( \frac{1}{4} - 1) [/tex] would be log of a negative number, so x = 2.

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