How do you solve #log_2 x+5=8-log_2 (x+7)#?

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Answer 1 · 2016-04-27T12:40:00

#x_1=-9,61#
#x_2=2,61#

#log_2x+5=8-log_2(x+7)#

#log_2x+log_2(x+7)=8-3#

#log_ab+log_a c=log_a(b*c)#

#log_2 x+log_2(x+7)=log_2 x(x+7)#

#log_2 x(x+7)=5#

#log_a b=k#

#b=a^k#

#x(x+7)=2^5#

#x^2+7x-25=0#

#Delta=sqrt(7^2+4*1*25)#

#Delta=sqrt(49+100)" "Delta=sqrt(149)" "Delta=12,21#

#x_1=(-7-12,21)/2" "x_1=-9,61#

#x_2=(-7+12,21)/2" "x_2=2,61#