What is the answer to log_{3}(x+2)+log_{3}(x+1)=log_{3}(x^2+1) ?

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

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log_{3}(x+2)+log_{3}(x+1)=log_{3}(x^2+1)

lo g_{3} (x + 2) + lo g_{3} (x + 1) = lo g_{3} (x^{2} + 1)

x = - \frac{1}{3}

Decimal Notation

x = - 0.33333 \dots

Solution steps

lo g_{3} (x + 2) + lo g_{3} (x + 1) = lo g_{3} (x^{2} + 1)

Apply log rules

(x + 2) (x + 1) = x^{2} + 1

Solve

(x + 2) (x + 1) = x^{2} + 1 : x = - \frac{1}{3}

x = - \frac{1}{3}

Verify Solutions:

x = - \frac{1}{3}

True

The solution is

x = - \frac{1}{3}

lo g_{22} (2 x) - 5 lo g_{2} (x) = 0

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

ln (8 - 2 x) = \frac{1}{2} ln (4 - x)

ln (x) = 3.47

lo g_{3} (2 x + 7) = 2

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