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

The answer to log_{2}(x)+log_{2}(x-4)=1 is x=2+sqrt(6)

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

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

x = 2 + 6

Decimal Notation

x = 4.44948 \dots

Solution steps

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

Apply log rules

x (x - 4) = 2

Solve

x (x - 4) = 2 : x = 2 + 6, x = 2 - 6

x = 2 + 6, x = 2 - 6

Verify Solutions:

x = 2 + 6

True

, x = 2 - 6

False

The solution is

x = 2 + 6

lo g_{10} (x) = 1 0^{6}

lo g_{x} (m) = n + 2

x, lo g_{x} (6 x + 16) = 2

2 ln (x + 2) - ln (x) = ln (2 x - 1)

lo g_{3} (x - 1) = - 1

What is the answer to log_{2}(x)+log_{2}(x-4)=1 ?
The answer to log_{2}(x)+log_{2}(x-4)=1 is x=2+sqrt(6)
Solve for x: log_{2}(x)+log_{2}(x-4)=1
The solution is x=2+sqrt(6)