What is the answer to log_{2}(x-1)+log_{2}(2x-1)=0 ?

The answer to log_{2}(x-1)+log_{2}(2x-1)=0 is x= 3/2

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log_{2}(x-1)+log_{2}(2x-1)=0

lo g_{2} (x - 1) + lo g_{2} (2 x - 1) = 0

x = \frac{3}{2}

Solution steps

lo g_{2} (x - 1) + lo g_{2} (2 x - 1) = 0

Apply log rules

(x - 1) (2 x - 1) = 1

Solve

(x - 1) (2 x - 1) = 1 : x = \frac{3}{2}, x = 0

x = \frac{3}{2}, x = 0

Verify Solutions:

x = \frac{3}{2}

True

, x = 0

False

The solution is

x = \frac{3}{2}

7.3 = lo g_{10} (x)

lo g_{6} (8 - 3 x) = 1

\frac{9}{2} lo g_{4} (x) = 7

\frac{9}{5} lo g_{4} (x) = - 2

\frac{9}{5} lo g_{4} (x) = - 6

What is the answer to log_{2}(x-1)+log_{2}(2x-1)=0 ?
The answer to log_{2}(x-1)+log_{2}(2x-1)=0 is x= 3/2
Solve for x: log_{2}(x-1)+log_{2}(2x-1)=0
The solution is x= 3/2