1.

Solve `(log)_2(3x-2)=(log)_(1/2)x`

Answer» `log_(2)(3x-2)=log_(1//2)x=-log_(2)x=log_(2)x^(-1)`
`or3x-2=x^(-1)or3x^(2)-2x=1`
`rArrx=1orx=-1//3`.
But `log_(2)(3x-2)` and `log_(1//2)x` are meaningful if `xgt2//3`. Hence, x=1


Discussion

No Comment Found