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

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

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

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

x = 1

Solution steps

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

Apply log rules

x (2 x + 1) = 3

Solve

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

x = 1, x = - \frac{3}{2}

Verify Solutions:

x = 1

True

, x = - \frac{3}{2}

False

The solution is

x = 1

lo g_{2} (x) = lo g_{4} (x^{2} + 2 x - 1)

lo g_{10} (10 x) - lo g_{10} (4 - x) = 2

lo g_{4} (y) + 2 = 3 \cdot lo g_{4} (4 y)

lo g_{2} (x^{2} - 4) - lo g_{2} (x - 2) = 3

lo g_{3} (4 x^{3} + 14 x^{2} + 13 x + 3) = lo g_{3} (x + 1)

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