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

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

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

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

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

No Solution for x \in R

Solution steps

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

Add

lo g_{2} (x + 2)

to both sides

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

Simplify

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

Apply log rules

x - 1 = 2 (x + 2)

Solve

x - 1 = 2 (x + 2) : x = - 5

x = - 5

Verify Solutions:

x = - 5

False

The solution is

No Solution for x \in R

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

lo g_{100} (x) = 1.5

lo g_{\frac{1}{4}} (x) = 0

lo g_{\frac{1}{4}} (x) = 1

ln (x + 7) - 2 ln (x + 1) = ln (16)

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