What is the answer to log_{2}(2x+1)-log_{2}(x+3)=0 ?

The answer to log_{2}(2x+1)-log_{2}(x+3)=0 is x=2

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log_{2}(2x+1)-log_{2}(x+3)=0

lo g_{2} (2 x + 1) - lo g_{2} (x + 3) = 0

x = 2

Solution steps

lo g_{2} (2 x + 1) - lo g_{2} (x + 3) = 0

Add

lo g_{2} (x + 3)

to both sides

lo g_{2} (2 x + 1) - lo g_{2} (x + 3) + lo g_{2} (x + 3) = 0 + lo g_{2} (x + 3)

Simplify

lo g_{2} (2 x + 1) = lo g_{2} (x + 3)

Apply log rules

2 x + 1 = x + 3

Solve

2 x + 1 = x + 3 : x = 2

x = 2

Verify Solutions:

x = 2

True

The solution is

x = 2

x, I n (lo g_{x} (3)) = 2

\frac{lo g _{4} ( 15 x + 1 ) 1}{2} = 2

lo g_{x} (243) = - 5

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

x, lo g_{3} (4 x - 1) = 3

What is the answer to log_{2}(2x+1)-log_{2}(x+3)=0 ?
The answer to log_{2}(2x+1)-log_{2}(x+3)=0 is x=2
Solve for x: log_{2}(2x+1)-log_{2}(x+3)=0
The solution is x=2