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

The answer to log_{2}(x+2)+log_{2}(x-4)=2 is x=1+sqrt(13)

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

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

x = 1 + 13

Decimal Notation

x = 4.60555 \dots

Solution steps

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

Apply log rules

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

Solve

(x + 2) (x - 4) = 4 : x = 1 + 13, x = 1 - 13

x = 1 + 13, x = 1 - 13

Verify Solutions:

x = 1 + 13

True

, x = 1 - 13

False

The solution is

x = 1 + 13

lo g_{2} (x + 2) + lo g_{2} (x - 4) = 3

lo g_{4} (x^{3} + 37) = 3

ln (x) = 12.39

lo g_{16} (x) = (lo g_{4} (x))

lo g_{2} (x^{2}) - lo g_{2} (5 x) = 4

What is the answer to log_{2}(x+2)+log_{2}(x-4)=2 ?
The answer to log_{2}(x+2)+log_{2}(x-4)=2 is x=1+sqrt(13)
Solve for x: log_{2}(x+2)+log_{2}(x-4)=2
The solution is x=1+sqrt(13)