How do you solve #2\log _ { 2} x = \log _ { 2} ( x - 2) + \log ( x + 4)#?

1 Answer
Purushottam M.
Apr 29, 2018

#2log_2x=log_2(x−2)+log_e(x+4)#
#2log_ex/log_e2=log_e(x-2)/log_e(2)+log_e(x+4)#
#2log_ex/log_e2=[log_e(x-2)+log_e(x+4)log_e2]/log_e2#
#2log_ex=log_e(x-2)+log_e(x+4)log_e2#
#log_ex^2-log_e(x-2)=log_e(x+4)log_e2#
#log_e(x^2/(x-2))=log_e(x+4)log_e2#
#ln(x^2/(x-2))=ln(x+4)ln2#
divide by#log_e2#

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