What is the answer to ln(2x+1)+ln(-x+1)=0 ?

The answer to ln(2x+1)+ln(-x+1)=0 is x=0,x= 1/2

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ln(2x+1)+ln(-x+1)=0

ln (2 x + 1) + ln (- x + 1) = 0

x = 0, x = \frac{1}{2}

Decimal Notation

x = 0, x = 0.5

Solution steps

ln (2 x + 1) + ln (- x + 1) = 0

Apply log rules

(2 x + 1) (- x + 1) = 1

Solve

(2 x + 1) (- x + 1) = 1 : x = 0, x = \frac{1}{2}

x = 0, x = \frac{1}{2}

Verify Solutions:

x = 0

True

, x = \frac{1}{2}

True

The solutions are

x = 0, x = \frac{1}{2}

lo g_{2} (x - 2) = 2 - lo g_{2} (x + 1)

x, ln (x) = ln (v) + c b

60 = 20 lo g_{10} (x)

lo g_{x + 3} (10 x + 9) = 2

lo g_{5} (x^{2} + 4 x) = 1

What is the answer to ln(2x+1)+ln(-x+1)=0 ?
The answer to ln(2x+1)+ln(-x+1)=0 is x=0,x= 1/2
Solve for x: ln(2x+1)+ln(-x+1)=0
The solution is x=0,x= 1/2