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

The answer to log_{2}(2x+3)+log_{2}(x-2)=2 is x= 5/2

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

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

x = \frac{5}{2}

Decimal Notation

x = 2.5

Solution steps

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

Apply log rules

(2 x + 3) (x - 2) = 4

Solve

(2 x + 3) (x - 2) = 4 : x = \frac{5}{2}, x = - 2

x = \frac{5}{2}, x = - 2

Verify Solutions:

x = \frac{5}{2}

True

, x = - 2

False

The solution is

x = \frac{5}{2}

4 ln (x - 3) = - 4

2 lo g_{3} (6) = lo g_{3} (x - 5)

lo g_{2} (2 p + 1) = lo g_{2} (5 p - 2)

ln (x^{2}) - ln (x) - ln (4) = 0

lo g_{20} (x) + lo g_{20} (x - 1) = 1

What is the answer to log_{2}(2x+3)+log_{2}(x-2)=2 ?
The answer to log_{2}(2x+3)+log_{2}(x-2)=2 is x= 5/2
Solve for x: log_{2}(2x+3)+log_{2}(x-2)=2
The solution is x= 5/2