What is the answer to log_{6}(3x)-log_{6}(2x)+4=0 ?

The answer to log_{6}(3x)-log_{6}(2x)+4=0 is No Solution for x\in\mathbb{R}

Solve for x: log_{6}(3x)-log_{6}(2x)+4=0

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

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lo g_{6} (3 x) - lo g_{6} (2 x) + 4 = 0

No Solution for x \in R

Solution steps

lo g_{6} (3 x) - lo g_{6} (2 x) + 4 = 0

Apply log rules

lo g_{6} (3) + lo g_{6} (x) - (lo g_{6} (2) + lo g_{6} (x)) + 4 = 0

Move

lo g_{6} (3)

to the right side

lo g_{6} (x) - (lo g_{6} (2) + lo g_{6} (x)) + 4 = - lo g_{6} (3)

Move

4

to the right side

lo g_{6} (x) - (lo g_{6} (2) + lo g_{6} (x)) = - lo g_{6} (3) - 4

Expand

lo g_{6} (x) - (lo g_{6} (2) + lo g_{6} (x)) : - lo g_{6} (2)

- lo g_{6} (2) = - lo g_{6} (3) - 4

The sides are not equal

No Solution for x \in R

3 ln (x - 6) = - 3

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

ln (t + 5) = 0

2 + 3 ln (x) = 12

5 lo g_{10} (n) = lo g_{2} (n)

What is the answer to log_{6}(3x)-log_{6}(2x)+4=0 ?
The answer to log_{6}(3x)-log_{6}(2x)+4=0 is No Solution for x\in\mathbb{R}
Solve for x: log_{6}(3x)-log_{6}(2x)+4=0
The solution is No Solution for x\in\mathbb{R}