Question

How do you solve `Log_2 X + Log_2 (X-3) = Log_2 (X^2-12)`?

Answer 1

Use the `log` multiplication rule to determine
`color(white)("XXX")x=4`

Explanation:

Log Multiplication Rule:
`color(white)("XXX")log_b(a)+log_b(c)= log_b(a*c)`

Therefore
`color(white)("XXX")log_2(x)+log_2(x-3)=log_2(x*(x-3)) = log_2(x^2-3x)`

and the given equation:
`color(white)("XXX")log_2(x)+log_2(x-3) = log_2(x^2-12)`
is equivalent to
`color(white)("XXX")log_2(x^2-3x) = log_2(x^2-12)`

This implies
`color(white)("XXX")x^2-3x = x^2-12`

`color(white)("XXX")-3x=-12`

`color(white)("XXX")x=4`

Answer 2

`x=4`

Explanation:

First of all, use the fact that for any base, the sum of two logarithms is the logarithm of the product of the arguments. In formulas:

`log(x)+log(y)=log(xy)`

This means that, in your case, we have that

`log_2(x)+log_2(x-3)=log_2(x(x-3))=log_2(x^2-3x)`

So, the equation becomes

`log_2(x^2-3x)=log_2(x^2-12)`

Now, the logarithm is an injective function. This means that two different numbers can't have the same logarithm. Again, in formulas, this means that

`log(a)=log(b) \iff a=b`

In your case,

`log_2(x^2-3x)=log_2(x^2-12) \iff x^2-3x=x^2-12`

Which is an equation that we can solve much easily:

`cancel(x^2)-3x=cancel(x^2)-12 \implies 3x=12 \implies x=4`