How do you solve #Log_2 (x-4) + log_2 (x+4) = 3#?

1 Answer
Dec 27, 2015

Answer:

Use sum of the logarithms rule and condense the logarithms on the left. Covert logarithms to exponent form to solve. The steps are given below.

Explanation:

#log_2(x-4)+log_2(x+4)=3#

Let us use the rule #log(A) + log(B) = log(AB)#

#log_2(x-4)(x+4)=3#

#log_2(x^2-16)=3# Note : #(a-b)(a+b) = a^2 - b^2#

Converting the logarithms to exponent form

If #log_b(a) =k# then #a=b^k#

#log_2(x^2-16) = 3#
# x^2-16 = 2^3#
#x^2-16 = 8#
#x^2-16+16 = 8+16#
#x^2=24#
take square root on both the sides.
#x = +-sqrt(24)#
#x=+-sqrt(4*6)#
#x=+-sqrt(4)*sqrt(6)#
#x=+-2sqrt(6)#

#x=2sqrt(6) or x = -2sqrt(6)#

Check the validity of solution by substituting the solution in the original equation.

We can see for #log_2(x-4)# if we substitute #x=-2sqrt(6)# the logarithm is not defined.

Therefore the final answer is #x=2sqrt(6)#