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

The answer to log_{2}(x+2)-log_{2}(3x-1)=-1 is x=5

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

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

x = 5

Solution steps

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

Add

lo g_{2} (3 x - 1)

to both sides

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

Simplify

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

Apply log rules

(x + 2) \cdot 2 = 3 x - 1

Solve

(x + 2) \cdot 2 = 3 x - 1 : x = 5

x = 5

Verify Solutions:

x = 5

True

The solution is

x = 5

ln (x) = - 2

lo g_{10} (4 x + 5) = 1 + lo g_{10} (x - 1)

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

lo g_{3} (x) = 5

ln (x) = \frac{1}{x}

What is the answer to log_{2}(x+2)-log_{2}(3x-1)=-1 ?
The answer to log_{2}(x+2)-log_{2}(3x-1)=-1 is x=5
Solve for x: log_{2}(x+2)-log_{2}(3x-1)=-1
The solution is x=5