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

The answer to log_{10}(x)+log_{10}(x+1)=1 is x=(-1+sqrt(41))/2

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

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

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

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

Decimal Notation

x = 2.70156 \dots

Solution steps

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

Apply log rules

x (x + 1) = 10

Solve

x (x + 1) = 10 : 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_{7} (x + 1) + lo g_{7} (x - 5) = 1

2 lo g_{9} (5 x) = 3

ln (3 x + 1) = 2

lo g_{10} (5 x) = lo g_{10} (2 x + 9)

ln ((x - 1)^{2} (x - 2)^{2}) = ln (4)

What is the answer to log_{10}(x)+log_{10}(x+1)=1 ?
The answer to log_{10}(x)+log_{10}(x+1)=1 is x=(-1+sqrt(41))/2
Solve for x: log_{10}(x)+log_{10}(x+1)=1
The solution is x=(-1+sqrt(41))/2