What is the answer to log_{2}(2x)-log_{2}(x+1)=0 ?

The answer to log_{2}(2x)-log_{2}(x+1)=0 is x=1

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log_{2}(2x)-log_{2}(x+1)=0

lo g_{2} (2 x) - lo g_{2} (x + 1) = 0

x = 1

Solution steps

lo g_{2} (2 x) - lo g_{2} (x + 1) = 0

Add

lo g_{2} (x + 1)

to both sides

lo g_{2} (2 x) - lo g_{2} (x + 1) + lo g_{2} (x + 1) = 0 + lo g_{2} (x + 1)

Simplify

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

Apply log rules

2 x = x + 1

Solve

2 x = x + 1 : x = 1

x = 1

Verify Solutions:

x = 1

True

The solution is

x = 1

lo g_{10} (x + 1) - lo g_{10} (x - 1) = 1

(lo g_{2} (x))^{2} + 2 \cdot lo g_{2} (x) - 2 = 0

(lo g_{10} (10 x)) 2 - 4 lo g_{10} (10 x) - 5 = 0

6 lo g_{10} (a) + 6 lo g_{10} (a - 9) = 2

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

What is the answer to log_{2}(2x)-log_{2}(x+1)=0 ?
The answer to log_{2}(2x)-log_{2}(x+1)=0 is x=1
Solve for x: log_{2}(2x)-log_{2}(x+1)=0
The solution is x=1