What is the answer to log_{2}(x+2)+log_{2}(x-4)=3 ?

The answer to log_{2}(x+2)+log_{2}(x-4)=3 is x=1+sqrt(17)

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log_{2}(x+2)+log_{2}(x-4)=3

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

x = 1 + 17

Decimal Notation

x = 5.12310 \dots

Solution steps

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

Apply log rules

(x + 2) (x - 4) = 8

Solve

(x + 2) (x - 4) = 8 : x = 1 + 17, x = 1 - 17

x = 1 + 17, x = 1 - 17

Verify Solutions:

x = 1 + 17

True

, x = 1 - 17

False

The solution is

x = 1 + 17

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

lo g_{3} (2) + lo g_{3} (x - 4) = 1

What is the answer to log_{2}(x+2)+log_{2}(x-4)=3 ?
The answer to log_{2}(x+2)+log_{2}(x-4)=3 is x=1+sqrt(17)
Solve for x: log_{2}(x+2)+log_{2}(x-4)=3
The solution is x=1+sqrt(17)