What is the answer to log_{6}(x)-log_{6}(x+1)=2 ?

The answer to log_{6}(x)-log_{6}(x+1)=2 is No Solution for x\in\mathbb{R}

Solve for x: log_{6}(x)-log_{6}(x+1)=2

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

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lo g_{6} (x) - lo g_{6} (x + 1) = 2

No Solution for x \in R

Solution steps

lo g_{6} (x) - lo g_{6} (x + 1) = 2

Add

lo g_{6} (x + 1)

to both sides

lo g_{6} (x) - lo g_{6} (x + 1) + lo g_{6} (x + 1) = 2 + lo g_{6} (x + 1)

Simplify

lo g_{6} (x) = 2 + lo g_{6} (x + 1)

Apply log rules

x = 36 (x + 1)

Solve

x = 36 (x + 1) : x = - \frac{36}{35}

x = - \frac{36}{35}

Verify Solutions:

x = - \frac{36}{35}

False

The solution is

No Solution for x \in R

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

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

5 ln (2 x + 3) + 6 = 16

5 lo g_{2} (x) = - 2

lo g_{10} (1 + x) = 0.599

What is the answer to log_{6}(x)-log_{6}(x+1)=2 ?
The answer to log_{6}(x)-log_{6}(x+1)=2 is No Solution for x\in\mathbb{R}
Solve for x: log_{6}(x)-log_{6}(x+1)=2
The solution is No Solution for x\in\mathbb{R}