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

The answer to log_{4}(x+2)-log_{4}(x-1)=0 is No Solution for x\in\mathbb{R}

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

The solution is No Solution for x\in\mathbb{R}

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

No Solution for x \in R

Solution steps

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

Add

lo g_{4} (x - 1)

to both sides

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

Simplify

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

Apply log rules

x + 2 = x - 1

Solve

x + 2 = x - 1 :

No Solution

No Solution for x \in R

x, ln (1 - x - y^{2}) = 0

ln (3 t + 1) - ln (2 t + 3) = 0

lo g_{3} (3 x - 12) - lo g_{3} (x) = 2

lo g_{10} (x) = 2.983

lo g_{6} (8 x) - 4 lo g_{6} (2) = 0

What is the answer to log_{4}(x+2)-log_{4}(x-1)=0 ?
The answer to log_{4}(x+2)-log_{4}(x-1)=0 is No Solution for x\in\mathbb{R}
Solve for x: log_{4}(x+2)-log_{4}(x-1)=0
The solution is No Solution for x\in\mathbb{R}