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

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

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

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

x = 1

Solution steps

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

Add

lo g_{2} (x)

to both sides

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

Simplify

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

Apply log rules

(x + 1) x = 2

Solve

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

x = 1, x = - 2

Verify Solutions:

x = 1

True

, x = - 2

False

The solution is

x = 1

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

lo g_{5} (x + 3) + lo g_{5} (x) = 2

lo g_{25} (x) = \frac{- 1}{2}

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

lo g_{10} (3 x) = lo g_{10} (9)

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