# Solve for x if log x^2 - log (x+2) = log 2

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### 3 Answers

You'll have to use the property that the difference of two logs is the same as the log of the ratio of the numbers:

Log(A/B) = Log A - log B

so,

log x^2 - log x+2 = log2

log( x^2/(x+2) ) = log 2

now, raise both sides to the tenth power

x^2/(x+2) = 2

x^2 = 2x + 4

x^2 -2x - 4 = 0

(x-1)^2-5 = 0

**x = 1 +/- sqrt(5)**

log x^2 - log (x+2) = log 2

using algorethim properties , we know that:

log a - log b = log a/b

==> log (x^2/ (x+2) = log 2

Now since the logarithm equals , then the bases are equals:

==> x^2 / (x+2) = 2

Cross multiply:

==> x^2 = 2(x+2)

==> x^2 = 2x - 4

==> x^2 - 2x + 4 = 0

==> x1= (2 + sqrt(4- 16) /2

= (2+ sqrt14 *i )/2

**Then, there are no solution for the equation because the algorethim is not defined for complex numbers.**

To solve logx^2-log(x+2) = log2:

Solution:

We lnow that loga - logb = loga/b.

Therefore lox^2-log(x+2) = log{x^2/(x+2)}

Therefore the given equation now becomes:

log(x^2/(x+2) = log2.

We take the antilogarothms of both sides:

x^2/(x+2) = 2.

We multiply both sides by (x+2):

x^2 = 2(x+2).

x^2 = 2x+4

Subtracting 2x-1 from both sides, we get:

x^2 -2x+1 = 5.

(x-1)^2 = 5.

We take the square root of both sides:

x-1 = +sqrt5, or x-1 = -sqrt5.

**x = 1+sqrt5,**

**Or x= 1-sqrt5.**