What is the answer to log_{6}(2x)-log_{6}(x+1)=0 ?

The answer to log_{6}(2x)-log_{6}(x+1)=0 is x=1

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log_{6}(2x)-log_{6}(x+1)=0

lo g_{6} (2 x) - lo g_{6} (x + 1) = 0

x = 1

Solution steps

lo g_{6} (2 x) - lo g_{6} (x + 1) = 0

Add

lo g_{6} (x + 1)

to both sides

lo g_{6} (2 x) - lo g_{6} (x + 1) + lo g_{6} (x + 1) = 0 + lo g_{6} (x + 1)

Simplify

lo g_{6} (2 x) = lo g_{6} (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

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

ln (x) = ln (x)

y, ln (y) = ln (x) + c

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

lo g_{3} (x + 2) = 5

What is the answer to log_{6}(2x)-log_{6}(x+1)=0 ?
The answer to log_{6}(2x)-log_{6}(x+1)=0 is x=1
Solve for x: log_{6}(2x)-log_{6}(x+1)=0
The solution is x=1