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

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

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

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

x = 2

Solution steps

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

Add

lo g_{4} (x - 1)

to both sides

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

Simplify

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

Apply log rules

(x + 2) (x - 1) = 4

Solve

(x + 2) (x - 1) = 4 : x = 2, x = - 3

x = 2, x = - 3

Verify Solutions:

x = 2

True

, x = - 3

False

The solution is

x = 2

x, ln (5 x - 3) = ln (2 x)

lo g_{2} (3 x + 1) + lo g_{2} (x - 3) = 5

2 = lo g_{2} (a^{2})

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

lo g_{2} (x) = 0.05

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