How do you solve #log_ 4 (x – 6) + log _ 4 x = 2#?

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Answer 1 · 2016-01-13T04:56:16

Step by step explanation is given below.

#log_4(x-6)+log_4x=2#

Step 1: Condense the left-hand side using rule
#log_b(P) + log_b(Q) = log_b(PQ)#

Step 2: Convert logarithms to exponent form using the rule
#log_b(a) = k -> a=b^k#

Now let us solve our problem

#log_4(x-6)+log_4(x)=2#

Step 1: Condensing...

#log_4(x-6)x=2#
#log_4(x^2-6x)=2#

Step 2: Converting to exponent form.

#x^2-6x = 2^4#

#=>x^2-6x = 16#

#=>x^2-6x-16=0#

#=>(x-8)(x+2)=0#

#x-8=0# or #x+2=0#

#x=8 or x = -2#

#x=-2# is not possible as #log_4(-2)# is not defined.

The solution is #x=8#