# Which is the only solution to the equation log_{3}(x^{2} + 6x) = log_{3}(2x + 12)?

x = -6, x = -2, x = 0, x = 2, x = 6

**Solution:**

It is given that expression is

log_{3}(x^{2} + 6x) = log_{3}(2x + 12)

Let us make use of the logarithmic rule

log_{b} x = log_{b} y

So x = y

x^{2} + 6x = 2x + 12

By subtracting 2x on both sides

x^{2} + 6x - 2x = 2x + 12 - 2x

x^{2} + 4x = 12

It can be written as

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

x^{2} + 6x - 2x - 12 = 0

Taking out the common terms

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

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

So we get

x + 6 = 0 and x - 2 = 0

x = - 6 and x = 2

As x = -6 does not satisfy the equation, x = 2 is the only solution.

Therefore, the only solution is x = 2.

## Which is the only solution to the equation log_{3}(x^{2} + 6x) = log_{3}(2x + 12)?

**Summary:**

The only solution to the equation log_{3}(x^{2} + 6x) = log_{3}(2x + 12) is x = 2.

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