What is the answer to ln(x^2+7x+12)-ln(x+4)=0 ?

The answer to ln(x^2+7x+12)-ln(x+4)=0 is x=-2

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ln(x^2+7x+12)-ln(x+4)=0

ln (x^{2} + 7 x + 12) - ln (x + 4) = 0

x = - 2

Solution steps

ln (x^{2} + 7 x + 12) - ln (x + 4) = 0

Add

ln (x + 4)

to both sides

ln (x^{2} + 7 x + 12) - ln (x + 4) + ln (x + 4) = 0 + ln (x + 4)

Simplify

ln (x^{2} + 7 x + 12) = ln (x + 4)

Apply log rules

x^{2} + 7 x + 12 = x + 4

Solve

x^{2} + 7 x + 12 = x + 4 : x = - 2, x = - 4

x = - 2, x = - 4

Verify Solutions:

x = - 2

True

, x = - 4

False

The solution is

x = - 2

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

- lo g_{10} (x) = 3.2

x, z = ln (2 x + 4 y)

ln (\frac{1}{x}) = 4

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

What is the answer to ln(x^2+7x+12)-ln(x+4)=0 ?
The answer to ln(x^2+7x+12)-ln(x+4)=0 is x=-2
Solve for x: ln(x^2+7x+12)-ln(x+4)=0
The solution is x=-2