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

2 Answers
Dec 16, 2015

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#

Dec 16, 2015

#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#