Solve log?x+ log(x+4)=log?12

log(x) + log(x + 4) = log(12)

may be rewritten as:

x(x + 4) = 12

x

^{2}+ 4x - 12 = 0(x - 2)(x + 6) = 0

x = 2 and x = -6

but x = -6 is considered a "non real" result, so we are left with

**x = 2**

Solve log?x+ log(x+4)=log?12

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Aliquippa, PA

log(x) + log(x + 4) = log(12)

may be rewritten as:

x(x + 4) = 12

x^{2} + 4x - 12 = 0

(x - 2)(x + 6) = 0

x = 2 and x = -6

but x = -6 is considered a "non real" result, so we are left with

Fort Lauderdale, FL

The first thing to know is properties of logs. If you recall the addition and subtraction properties:

Addition:

log(a) + log(b) = log(a*b)

Subtraction:

log(a) - log(b) = log(a/b)

Then you will find this a little easier.

Furthermore, if you can express the left and right sides as logarithms, then you may solve. Keep in mind that you cannot take the log of a negative number. So if you left side is an expression and the right side is a negative number, there is no solution. Example:

log(3x)+log(x+1) = log(-3)

There is no solution to this. Even though you may go through the process of combining logs and end up with an answer, it cannot be validated by the original equation.

Getting back to the problem:

log(x)+log(x+4)=log(12)

Combine the left side of the equation:

log(x*(x+4))=log(12)

So,

x(x+4)=12

x^{2}+4x=12

x^{2}+4x-12=0

Solve by quadratic formula or factoring:

(x+6)(x-2)=0

The solutions are x={-6,2}

You must plug these into the original equation to make sure they are validated. Not all solutions of x will work in the original equation. So Checking:

For x=-6:

log(-6)+log(-6+4)=log(12)

You cannot take the log of a negative number, so: no solution here.

Then

x=2;

log(2)+log(2+4)=log(12)

log(2)+log(6)=log(12)

log(12)=log(12)

This is the answer.

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