What is the answer to 2log_{x}(3x-1)=log_{x}(4x) ?

The answer to 2log_{x}(3x-1)=log_{x}(4x) is No Solution for x\in\mathbb{R}

Solve for x: 2log_{x}(3x-1)=log_{x}(4x)

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

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2 lo g_{x} (3 x - 1) = lo g_{x} (4 x)

No Solution for x \in R

Solution steps

2 lo g_{x} (3 x - 1) = lo g_{x} (4 x)

Apply log rules

(3 x - 1)^{2} = 4 x

Solve

(3 x - 1)^{2} = 4 x : x = 1, x = \frac{1}{9}

x = 1, x = \frac{1}{9}

Verify Solutions:

x = 1

False

, x = \frac{1}{9}

False

The solution is

No Solution for x \in R

lo g_{x} (\frac{1}{2}) = - 2

x, lo g_{2} (6 x + 3) = 5

lo g_{x + 5} (4 x) = 3

lo g_{3} (a) = 2

ln (3 x) - ln (x - 4) = ln (x + 3)

What is the answer to 2log_{x}(3x-1)=log_{x}(4x) ?
The answer to 2log_{x}(3x-1)=log_{x}(4x) is No Solution for x\in\mathbb{R}
Solve for x: 2log_{x}(3x-1)=log_{x}(4x)
The solution is No Solution for x\in\mathbb{R}