What is the answer to log_{10}(x)+log_{10}(x+1)=2 ?

The answer to log_{10}(x)+log_{10}(x+1)=2 is x=(-1+sqrt(401))/2

Solve for x: log_{10}(x)+log_{10}(x+1)=2

The solution is x=(-1+sqrt(401))/2

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

x = \frac{- 1 + 401}{2}

Decimal Notation

x = 9.51249 \dots

Solution steps

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

Apply log rules

x (x + 1) = 100

Solve

x (x + 1) = 100 : x = \frac{- 1 + 401}{2}, x = \frac{- 1 - 401}{2}

x = \frac{- 1 + 401}{2}, x = \frac{- 1 - 401}{2}

Verify Solutions:

x = \frac{- 1 + 401}{2}

True

, x = \frac{- 1 - 401}{2}

False

The solution is

x = \frac{- 1 + 401}{2}

lo g_{10} (2 x - 3) = 1

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

lo g_{10} ((5 x - 3)^{2}) + lo g_{10} ((2 x + 3)^{3}) = 2

ln (x + 1) = 2 + ln (x - 1)

lo g_{2} (y) - 8 \cdot lo g_{y} (2) = 2

What is the answer to log_{10}(x)+log_{10}(x+1)=2 ?
The answer to log_{10}(x)+log_{10}(x+1)=2 is x=(-1+sqrt(401))/2
Solve for x: log_{10}(x)+log_{10}(x+1)=2
The solution is x=(-1+sqrt(401))/2