What is the answer to log_{e}(x)+log_{e}(7x+1)=1 ?

The answer to log_{e}(x)+log_{e}(7x+1)=1 is x=(-1+sqrt(1+28e))/(14)

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log_{e}(x)+log_{e}(7x+1)=1

lo g_{e} (x) + lo g_{e} (7 x + 1) = 1

x = \frac{- 1 + 1 + 28 e}{14}

Decimal Notation

x = 0.55580 \dots

Solution steps

lo g_{e} (x) + lo g_{e} (7 x + 1) = 1

Apply log rules

x (7 x + 1) = e

Solve

x (7 x + 1) = e : x = \frac{- 1 + 1 + 28 e}{14}, x = \frac{- 1 - 1 + 28 e}{14}

x = \frac{- 1 + 1 + 28 e}{14}, x = \frac{- 1 - 1 + 28 e}{14}

Verify Solutions:

x = \frac{- 1 + 1 + 28 e}{14}

True

, x = \frac{- 1 - 1 + 28 e}{14}

False

The solution is

x = \frac{- 1 + 1 + 28 e}{14}

ln (8^{x^{2}}) = ln (9^{x})

4 ln (x) = 2 ln (16)

lo g_{x} (243) = \frac{5}{3}

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

25 = 19.725 ln (x) - 114.69

What is the answer to log_{e}(x)+log_{e}(7x+1)=1 ?
The answer to log_{e}(x)+log_{e}(7x+1)=1 is x=(-1+sqrt(1+28e))/(14)