How do you solve #log_2(x+1)-log_4(x^2+2x-1)=1/2#?

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Answer 1 · 2016-08-15T18:47:52

#x=1#

#log_2(x+1)-log_4(x^2+2x-1)=1/2#

#hArrlog_4((x+1)^2)-log_4(x^2+2x-1)=1/2#

#hArrlog_4((x^2+2x+1)/(2(x^2+2x-1)))=1/2#

#hArr(x^2+2x+1)/(x^2+2x-1))=4^(1/2)=2# or

#x^2+2x+1=2x^2+4x-2# or

#x^2+2x-3=0# or

#x=(-2+-sqrt(2^2-4×1×(-3)))/2#

= #(-2+-sqrt(4+12))/2#

= #(-2+-sqrt16)/2#

= #(-1+-2)=-3# or #+1#

But as #-3# is not permissible,

#x=1#

Answer 2 · 2016-08-15T19:09:58

#x=1# is a soln., while #x=-3# is extraneous.

We will use the Change of base Rule for Log # : log_b c=log_a c/log_ab#

Now, #log_2 (x+1)-log_4(x^2+2x-1)=1/2#.

#rArr log_e (x+1)/log_e 2-log_e(x^2+2x-1)/log_e4=1/2#

#rArr log_e(x+1)/log_e2-log_e(x^2+2x-1)/log_e2^2=1/2#

#rArr ln(x+1)/ln2-ln(x^2+2x-1)/(2ln2)=1/2#

#rArr (2ln(x+1))/(2ln2)-ln(x^2+2x-1)/(2ln2)=1/2#

#rArr (2ln(x+1))-ln(x^2+2x-1)=1/2*2ln2#

#rArr ln(x+1)^2-ln(x^2+2x-1)=ln2#

#rArr ln{(x+1)^2/(x^2+2x-1)}=ln2#

Since, #ln# is #1-1# function, wehave,

#(x+1)^2/(x^2+2x-1)=2#

#:. x^2+2x+1=2x^2+4x-2#

#:. x^2+2x-3=0#.

#:. (x-1)(x+3)=0#

#:. x=1, or, x=-3#

Since, #x=-3# makes #log_2(x+1)# undefined, it is an extraneous root.

Hence, #x=1# is a soln., while #x=-3# is extraneous.