How do you solve 2\log _ { 2} x = \log _ { 2} ( x - 2) + \log ( x + 4)2log2x=log2(x2)+log(x+4)?

1 Answer
Apr 29, 2018

2log_2x=log_2(x−2)+log_e(x+4)2log2x=log2(x2)+loge(x+4)
2log_ex/log_e2=log_e(x-2)/log_e(2)+log_e(x+4)2logexloge2=loge(x2)loge(2)+loge(x+4)
2log_ex/log_e2=[log_e(x-2)+log_e(x+4)log_e2]/log_e22logexloge2=loge(x2)+loge(x+4)loge2loge2
2log_ex=log_e(x-2)+log_e(x+4)log_e22logex=loge(x2)+loge(x+4)loge2
log_ex^2-log_e(x-2)=log_e(x+4)log_e2logex2loge(x2)=loge(x+4)loge2
log_e(x^2/(x-2))=log_e(x+4)log_e2loge(x2x2)=loge(x+4)loge2
ln(x^2/(x-2))=ln(x+4)ln2ln(x2x2)=ln(x+4)ln2
divide bylog_e2loge2