What is the answer to ln(x^2+6)-ln(x-2)=ln(x+1) ?

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

Solve for x: ln(x^2+6)-ln(x-2)=ln(x+1)

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

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ln (x^{2} + 6) - ln (x - 2) = ln (x + 1)

No Solution for x \in R

Solution steps

ln (x^{2} + 6) - ln (x - 2) = ln (x + 1)

Add

ln (x - 2)

to both sides

ln (x^{2} + 6) - ln (x - 2) + ln (x - 2) = ln (x + 1) + ln (x - 2)

Simplify

ln (x^{2} + 6) = ln (x + 1) + ln (x - 2)

Apply log rules

x^{2} + 6 = (x + 1) (x - 2)

Solve

x^{2} + 6 = (x + 1) (x - 2) : x = - 8

x = - 8

Verify Solutions:

x = - 8

False

The solution is

No Solution for x \in R

lo g_{10} (n) = 1.0043

lo g_{10} (x) + lo g_{10} (7) = lo g_{10} (4)

lo g_{\frac{1}{2}} (x^{2} - 3) = lo g_{2} (x^{2})

u, lo g_{u} (\frac{15}{16}) = v

ln (x) = \frac{1}{2}

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