Question

For the love of God help me !! I'm desperate for it tomorrow

Answer 1

`\log_2(5x - 2) - \log_2 x - \log_2(x - 1) = 2`

`\log_2 ( (5x - 2)/(x(x-1)) ) = 2`

`\log_2 ( (5x - 2)/(x(x-1)) ) = \log_2 4`

`(5x - 2)/(x(x-1)) = 4`

`5x - 2 = 4x(x - 1)`

`5x - 2 = 4x^2 - 4x`

`0 = 4x^2 - 9x + 2`

`0 = (4x - 1)(x - 2)`

`\implies x = 2 \text{ or } x = (1)/(4)`

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

Answer 2

`D:5x-2>0 \wedge x>0 \wedge x-1>0\\D:5x>2 \wedge x>0 \wedge x>1\\D: x>(2)/(5) \wedge x>1\\D:x>1\\\log_2(5x-2)-\log_2x-\log_2(x-1)=2\\\log_2(5x-2)/(x(x-1))=\log_24\\(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=(1)/(4)\\(1)/(4)\\ot \in D \Rightarrow \boxed{x=2}`