What is the answer to log_{2}(x+1)+log_{2}(x-5)=4 ?

The answer to log_{2}(x+1)+log_{2}(x-5)=4 is x=7

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log_{2}(x+1)+log_{2}(x-5)=4

lo g_{2} (x + 1) + lo g_{2} (x - 5) = 4

x = 7

Solution steps

lo g_{2} (x + 1) + lo g_{2} (x - 5) = 4

Apply log rules

(x + 1) (x - 5) = 16

Solve

(x + 1) (x - 5) = 16 : x = 7, x = - 3

x = 7, x = - 3

Verify Solutions:

x = 7

True

, x = - 3

False

The solution is

x = 7

x, ln (2 x) = 4

lo g_{3} (x) = lo g_{3} (12)

lo g_{10} (x) = 3.2614400000000 \dots

2 ln (3 x - 5) - 2 = 2

(lo g_{3} (x) - 2) lo g_{3} (x) = 0

What is the answer to log_{2}(x+1)+log_{2}(x-5)=4 ?
The answer to log_{2}(x+1)+log_{2}(x-5)=4 is x=7
Solve for x: log_{2}(x+1)+log_{2}(x-5)=4
The solution is x=7