What is the answer to log_{3}(2-3x)=log_{9}(6x^2-19x+2) ?

The answer to log_{3}(2-3x)=log_{9}(6x^2-19x+2) is x=-1/3 ,x=-2

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log_{3}(2-3x)=log_{9}(6x^2-19x+2)

lo g_{3} (2 - 3 x) = lo g_{9} (6 x^{2} - 19 x + 2)

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

Decimal Notation

x = - 0.33333 \dots, x = - 2

Solution steps

lo g_{3} (2 - 3 x) = lo g_{9} (6 x^{2} - 19 x + 2)

Apply log rules

(2 - 3 x)^{2} = 6 x^{2} - 19 x + 2

Solve

(2 - 3 x)^{2} = 6 x^{2} - 19 x + 2 : x = - \frac{1}{3}, x = - 2

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

Verify Solutions:

x = - \frac{1}{3}

True

, x = - 2

True

The solutions are

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

5 = lo g_{10} (x)

lo g_{10} (3 x + 5) + 1 = lo g_{10} (2 x + 1)

lo g_{3} (3^{x} - 8) = 2 - x

lo g_{10} (x^{3}) = \frac{1}{2}

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

What is the answer to log_{3}(2-3x)=log_{9}(6x^2-19x+2) ?
The answer to log_{3}(2-3x)=log_{9}(6x^2-19x+2) is x=-1/3 ,x=-2
Solve for x: log_{3}(2-3x)=log_{9}(6x^2-19x+2)
The solution is x=-1/3 ,x=-2