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

The answer to log_{3}(3x-1)-log_{3}(2x+3)=1+log_{3}(x) is No Solution for x\in\mathbb{R}

Solve for x: log_{3}(3x-1)-log_{3}(2x+3)=1+log_{3}(x)

The solution is No Solution for x\in\mathbb{R}

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

No Solution for x \in R

Solution steps

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

Add

lo g_{3} (2 x + 3)

to both sides

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

Simplify

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

Apply log rules

3 x - 1 = 3 x (2 x + 3)

Solve

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

x = \frac{- 3 + 3}{6}, x = - \frac{3 + 1}{2 3}

Verify Solutions:

x = \frac{- 3 + 3}{6}

False

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

False

The solution is

No Solution for x \in R

lo g_{3} (x) + lo g_{3} (x + 6) = 3

2 lo g_{10} (x) - lo g_{10} (x - 6) = 0

4 lo g_{6} (x) - 2 lo g_{10} (x^{6}) = 7

(- 3 x) lo g_{10} (x + 8) = 0

x, ln (\frac{x}{x _{0}}) = u (t - t_{0})

What is the answer to log_{3}(3x-1)-log_{3}(2x+3)=1+log_{3}(x) ?
The answer to log_{3}(3x-1)-log_{3}(2x+3)=1+log_{3}(x) is No Solution for x\in\mathbb{R}
Solve for x: log_{3}(3x-1)-log_{3}(2x+3)=1+log_{3}(x)
The solution is No Solution for x\in\mathbb{R}