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

The answer to log_{2}(x-2)+log_{2}(x+3)=2 is x=(-1+sqrt(41))/2

Solve for x: log_{2}(x-2)+log_{2}(x+3)=2

The solution is x=(-1+sqrt(41))/2

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

x = \frac{- 1 + 41}{2}

Decimal Notation

x = 2.70156 \dots

Solution steps

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

Apply log rules

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

Solve

(x - 2) (x + 3) = 4 : x = \frac{- 1 + 41}{2}, x = \frac{- 1 - 41}{2}

x = \frac{- 1 + 41}{2}, x = \frac{- 1 - 41}{2}

Verify Solutions:

x = \frac{- 1 + 41}{2}

True

, x = \frac{- 1 - 41}{2}

False

The solution is

x = \frac{- 1 + 41}{2}

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

lo g_{2} (N) = 3

x, ln (x) = - 7

lo g_{3} (3 x + 8) = 5

lo g_{2} (\frac{x}{32000}) = 0.5

What is the answer to log_{2}(x-2)+log_{2}(x+3)=2 ?
The answer to log_{2}(x-2)+log_{2}(x+3)=2 is x=(-1+sqrt(41))/2
Solve for x: log_{2}(x-2)+log_{2}(x+3)=2
The solution is x=(-1+sqrt(41))/2