Question

Log3(x2+2x-2×2/3)=-1 How?

Answer 1

`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!