Log3(x2+2x-2×2/3)=-1 How?

1 Answer
May 15, 2018

#x_1=-(3+2sqrt6)/3#
#x_2=(-3+2sqrt6)/3#

Explanation:

#Log_3(x²+2x-4/3)=-1#
#D_f:x²+2x-4/3>0#
#D_f:x in ]-oo;-(3+sqrt21)/3[U] ((-3+sqrt21)/3);+oo[#
Now :
#ln(x²+2x-4/3)/ln3=-1#
#ln(x²+2x-4/3)=-1*ln3#
Because #alnb=lnb^a#,
#ln(x²+2x-4/3)=ln(1/3)#
#x²+2x-4/3=1/3#
#x²+2x-5/3=0#
#Δ=2²-4*1*(-5/3)#
#Δ=4+20/3#
#Δ=32/3#
So: #x_1=-(2+sqrt(32/3))/2#
#x_2=(-2+sqrt(32/3))/2#
#x_1=-(3+2sqrt6)/3#
#x_2=(-3+2sqrt6)/3#
\0/ here's our answer!