What is the answer to log_{4}(2x+1)-log_{2}(1-x)=2 ?

The answer to log_{4}(2x+1)-log_{2}(1-x)=2 is x= 5/8

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log_{4}(2x+1)-log_{2}(1-x)=2

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

x = \frac{5}{8}

Decimal Notation

x = 0.625

Solution steps

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

Apply log rules

2 x + 1 = 16 (1 - x)^{2}

Solve

2 x + 1 = 16 (1 - x)^{2} : x = \frac{3}{2}, x = \frac{5}{8}

x = \frac{3}{2}, x = \frac{5}{8}

Verify Solutions:

x = \frac{3}{2}

False

, x = \frac{5}{8}

True

The solution is

x = \frac{5}{8}

lo g_{4} (8) + lo g_{4} (x - 4) = 3

2 \cdot lo g_{3} (x - 1) = 4

lo g_{2} (n) = 10000000

37.162 = lo g_{10} (x)

lo g_{10} (6 x - 3) = - 4

What is the answer to log_{4}(2x+1)-log_{2}(1-x)=2 ?
The answer to log_{4}(2x+1)-log_{2}(1-x)=2 is x= 5/8
Solve for x: log_{4}(2x+1)-log_{2}(1-x)=2
The solution is x= 5/8