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

The answer to log_{2}(x-2)+log_{2}(2x-1)=0 is x=(5+sqrt(17))/4

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

The solution is x=(5+sqrt(17))/4

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

x = \frac{5 + 17}{4}

Solution steps

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

Apply log rules

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

Solve

(x - 2) (2 x - 1) = 1 : x = \frac{5 + 17}{4}, x = \frac{5 - 17}{4}

x = \frac{5 + 17}{4}, x = \frac{5 - 17}{4}

Verify Solutions:

x = \frac{5 + 17}{4}

True

, x = \frac{5 - 17}{4}

False

The solution is

x = \frac{5 + 17}{4}

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

6 (ln (4 x) + 1) = 0

ln (5 - 2 x) = 3

2 lo g_{3} (x) = lo g_{10} (x^{2})

ln (3 ln (x)) = 0

What is the answer to log_{2}(x-2)+log_{2}(2x-1)=0 ?
The answer to log_{2}(x-2)+log_{2}(2x-1)=0 is x=(5+sqrt(17))/4
Solve for x: log_{2}(x-2)+log_{2}(2x-1)=0
The solution is x=(5+sqrt(17))/4