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

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

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log_{2}(k+2)-log_{2}(3-k)=-2

lo g_{2} (k + 2) - lo g_{2} (3 - k) = - 2

k = - 1

Solution steps

lo g_{2} (k + 2) - lo g_{2} (3 - k) = - 2

Add

lo g_{2} (3 - k)

to both sides

lo g_{2} (k + 2) - lo g_{2} (3 - k) + lo g_{2} (3 - k) = - 2 + lo g_{2} (3 - k)

Simplify

lo g_{2} (k + 2) = - 2 + lo g_{2} (3 - k)

Apply log rules

(k + 2) \cdot 4 = 3 - k

Solve

(k + 2) \cdot 4 = 3 - k : k = - 1

k = - 1

Verify Solutions:

k = - 1

True

The solution is

k = - 1

lo g_{2} (x) = 1 0^{6}

2 ln (x) + ln (x) = 0

ln (t) = 0

lo g_{7} (x - 2) + lo g_{7} (x + 4) = 1

3 ln (x - 3) + 4 = 5

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