What is the answer to log_{2}(x+19)+log_{2}(x+4)=4 ?

The answer to log_{2}(x+19)+log_{2}(x+4)=4 is x=-3

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log_{2}(x+19)+log_{2}(x+4)=4

lo g_{2} (x + 19) + lo g_{2} (x + 4) = 4

x = - 3

Solution steps

lo g_{2} (x + 19) + lo g_{2} (x + 4) = 4

Apply log rules

(x + 19) (x + 4) = 16

Solve

(x + 19) (x + 4) = 16 : x = - 3, x = - 20

x = - 3, x = - 20

Verify Solutions:

x = - 3

True

, x = - 20

False

The solution is

x = - 3

x^{1 + l n (x)} = e^{4} \cdot x

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

- 3 - ln (- 3 + x) = 0

a^{5} = lo g_{5} (x^{5})

lo g_{X} (16) = - 1

What is the answer to log_{2}(x+19)+log_{2}(x+4)=4 ?
The answer to log_{2}(x+19)+log_{2}(x+4)=4 is x=-3
Solve for x: log_{2}(x+19)+log_{2}(x+4)=4
The solution is x=-3