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

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

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

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

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

No Solution for x \in R

Solution steps

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

Add

lo g_{3} (x + 2)

to both sides

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

Simplify

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

Apply log rules

2 x + 3 = 81 (x + 2)

Solve

2 x + 3 = 81 (x + 2) : x = - \frac{159}{79}

x = - \frac{159}{79}

Verify Solutions:

x = - \frac{159}{79}

False

The solution is

No Solution for x \in R

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

lo g_{17} (x^{3} + 2 x^{2} - x + 15) = 1

lo g_{4} (22 - 4 x) = 2

lo g_{2} (x^{5}) = lo g_{10} (2 x)^{4}

lo g_{5} (x^{2}) - 4 = 2 + ∣ lo g_{5} (x) - 3 ∣

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