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

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

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

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

x = 2

Solution steps

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

Apply log rules

x (x - 1) = 2

Solve

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

x = 2, x = - 1

Verify Solutions:

x = 2

True

, x = - 1

False

The solution is

x = 2

lo g_{10} (x) + lo g_{10} (x + 9) = 1

lo g_{10} (x) - 1 = 0

y, ln (1 - 2 y) = t

lo g_{10} (x + 2) + lo g_{10} (x - 1) = 1

lo g_{10} (x) = 1

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