What is the answer to log_{6}(3-m)+log_{6}(-m-2)=1 ?

The answer to log_{6}(3-m)+log_{6}(-m-2)=1 is m=-3

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log_{6}(3-m)+log_{6}(-m-2)=1

lo g_{6} (3 - m) + lo g_{6} (- m - 2) = 1

m = - 3

Solution steps

lo g_{6} (3 - m) + lo g_{6} (- m - 2) = 1

Apply log rules

(3 - m) (- m - 2) = 6

Solve

(3 - m) (- m - 2) = 6 : m = 4, m = - 3

m = 4, m = - 3

Verify Solutions:

m = 4

False

, m = - 3

True

The solution is

m = - 3

x^{l n (4)} + 4^{l n (x)} = 32

ln (x - 3) - ln (x + 5) = 4

0.72 = ln (y)

ln (x) + ln (2) = 1

2 lo g_{10} (x - 2) = 4 - x

What is the answer to log_{6}(3-m)+log_{6}(-m-2)=1 ?
The answer to log_{6}(3-m)+log_{6}(-m-2)=1 is m=-3
Solve for m: log_{6}(3-m)+log_{6}(-m-2)=1
The solution is m=-3