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

The answer to log_{2}(2x+1)-log_{2}(x-1)=4 is x= 17/14

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

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

x = \frac{17}{14}

Decimal Notation

x = 1.21428 \dots

Solution steps

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

Add

lo g_{2} (x - 1)

to both sides

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

Simplify

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

Apply log rules

2 x + 1 = 16 (x - 1)

Solve

2 x + 1 = 16 (x - 1) : x = \frac{17}{14}

x = \frac{17}{14}

Verify Solutions:

x = \frac{17}{14}

True

The solution is

x = \frac{17}{14}

lo g_{5} (2 x - 1) = 3

x = lo g_{10} (x)

ln (5 x - 1) - ln (x) = 3

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

lo g_{10} (6 x) + ln (5 x) = 0

What is the answer to log_{2}(2x+1)-log_{2}(x-1)=4 ?
The answer to log_{2}(2x+1)-log_{2}(x-1)=4 is x= 17/14
Solve for x: log_{2}(2x+1)-log_{2}(x-1)=4
The solution is x= 17/14