What is the answer to log_{2}(11-x)=log_{2}(x+1)+3 ?

The answer to log_{2}(11-x)=log_{2}(x+1)+3 is x= 1/3

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log_{2}(11-x)=log_{2}(x+1)+3

lo g_{2} (11 - x) = lo g_{2} (x + 1) + 3

x = \frac{1}{3}

Solution steps

lo g_{2} (11 - x) = lo g_{2} (x + 1) + 3

Apply log rules

11 - x = (x + 1) \cdot 8

Solve

11 - x = (x + 1) \cdot 8 : x = \frac{1}{3}

x = \frac{1}{3}

Verify Solutions:

x = \frac{1}{3}

True

The solution is

x = \frac{1}{3}

lo g_{a} (N) = ln (x)

lo g_{3} (2 x) = 1

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

x, ln (x + y - 5) = 0

lo g_{4} (x) - lo g_{8} (x) = 1

What is the answer to log_{2}(11-x)=log_{2}(x+1)+3 ?
The answer to log_{2}(11-x)=log_{2}(x+1)+3 is x= 1/3
Solve for x: log_{2}(11-x)=log_{2}(x+1)+3
The solution is x= 1/3