Help me, please, i"m desperate
Answer (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
The problem involves solving an inequality and logarithmic equation step by step. The valid solution after determining the domain and solving is x = 2, as it satisfies all conditions. Hence, the solution is x = 2.
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