for the love of God help me !! I'm desperate for it tomorrow
Answer (3)
lo g 2 ( 5 x − 2 ) − lo g 2 x − lo g 2 ( x − 1 ) = 2 lo g 2 ( x ( x − 1 ) 5 x − 2 ) = 2 lo g 2 ( x ( x − 1 ) 5 x − 2 ) = lo g 2 4 x ( x − 1 ) 5 x − 2 = 4 5 x − 2 = 4 x ( x − 1 ) 5 x − 2 = 4 x 2 − 4 x 0 = 4 x 2 − 9 x + 2 0 = ( 4 x − 1 ) ( x − 2 ) ⟹ x = 2 or x = 4 1
By putting one quarter into the equation we can see that it cannot work in any way involving real numbers as lo g 2 ( 4 1 − 1 ) would be log of a negative number, so x = 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}"> D : 5 x − 2 > 0 ∧ x > 0 ∧ x − 1 > 0 D : 5 x > 2 ∧ x > 0 ∧ x > 1 D : x > 5 2 ∧ x > 1 D : x > 1 lo g 2 ( 5 x − 2 ) − lo g 2 x − lo g 2 ( x − 1 ) = 2 lo g 2 x ( x − 1 ) 5 x − 2 = lo g 2 4 x ( x − 1 ) 5 x − 2 = 4 4 x ( x − 1 ) = 5 x − 2 4 x 2 − 4 x = 5 x − 2 4 x 2 − 9 x + 2 = 0 4 x 2 − x − 8 x + 2 = 0 x ( 4 x − 1 ) − 2 ( 4 x − 1 ) = 0 ( x − 2 ) ( 4 x − 1 ) = 0 x = 2 ∨ x = 4 1 4 1 ∈ D ⇒ x = 2
To solve lo g 2 ( 5 x − 2 ) − lo g 2 x − lo g 2 ( x − 1 ) = 2 , we combine logarithms and set up a quadratic equation. The valid solution to the equation is x = 2 , as the other potential solution leads to an invalid logarithm. Thus, x = 2 is the final answer.
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