please help me I'm desperate.

lo g ( 4 x − 1 ) − lo g ( x + 2 ) = lo g x lo g ( x + 2 4 x − 1 ​ ) = lo g x x + 2 4 x − 1 ​ = x 4 x − 1 = x ( x + 2 ) 4 x − 1 = x 2 + 2 x 0 = x 2 − 2 x + 1 0 = ( x − 1 ) 2 ⇒ x = 1

Answered by superkoopasirf | 2024-06-10

0 \wedge x+2 >0 \wedge x>0\\ D:4x>1 \wedge x>-2 \wedge x>0\\ D:x>\frac{1}{4} \wedge x>0\\ D:x>\frac{1}{4}\\\\ \log(4x-1)-\log(x+2)=\log x\\ \log\frac{4x-1}{x+2}=\log x\\ \frac{4x-1}{x+2}=x\\ x(x+2)=4x-1\\ x^2+2x=4x-1\\ x^2-2x+1=0\\ (x-1)^2=0\\ \boxed{x=1}"> D : 4 x − 1 > 0 ∧ x + 2 > 0 ∧ x > 0 D : 4 x > 1 ∧ x > − 2 ∧ x > 0 D : x > 4 1 ​ ∧ x > 0 D : x > 4 1 ​ lo g ( 4 x − 1 ) − lo g ( x + 2 ) = lo g x lo g x + 2 4 x − 1 ​ = lo g x x + 2 4 x − 1 ​ = x x ( x + 2 ) = 4 x − 1 x 2 + 2 x = 4 x − 1 x 2 − 2 x + 1 = 0 ( x − 1 ) 2 = 0 x = 1 ​

Answered by konrad509 | 2024-06-10

The equation lo g ( 4 x − 1 ) − lo g ( x + 2 ) = lo g x simplifies to x = 1 after applying logarithmic properties and solving a quadratic equation. This solution is valid as it keeps all logarithmic expressions defined. Therefore, the solution to the equation is x = 1 .
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Answered by superkoopasirf | 2024-12-26