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

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

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

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

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

No Solution for x \in R

Solution steps

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

Add

lo g_{3} (x - 1)

to both sides

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

Simplify

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

Apply log rules

108 x = x - 1

Solve

108 x = x - 1 : x = - \frac{1}{107}

x = - \frac{1}{107}

Verify Solutions:

x = - \frac{1}{107}

False

The solution is

No Solution for x \in R

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

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

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

x, - ln (y) = ln (x) + c

lo g_{4} (9 x - 4) = 2

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