What is the answer to log_{4}(2x)-log_{4}(x+1)=0 ?

The answer to log_{4}(2x)-log_{4}(x+1)=0 is x=1

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log_{4}(2x)-log_{4}(x+1)=0

lo g_{4} (2 x) - lo g_{4} (x + 1) = 0

x = 1

Solution steps

lo g_{4} (2 x) - lo g_{4} (x + 1) = 0

Add

lo g_{4} (x + 1)

to both sides

lo g_{4} (2 x) - lo g_{4} (x + 1) + lo g_{4} (x + 1) = 0 + lo g_{4} (x + 1)

Simplify

lo g_{4} (2 x) = lo g_{4} (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_{x} (1) = 35

0 = 2 lo g_{e} (\frac{1}{x})

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

ln (x) = 2.37

- 10 = 20 lo g_{10} (\frac{2}{x})

What is the answer to log_{4}(2x)-log_{4}(x+1)=0 ?
The answer to log_{4}(2x)-log_{4}(x+1)=0 is x=1
Solve for x: log_{4}(2x)-log_{4}(x+1)=0
The solution is x=1