Giải phương trình loga (tìm x)

ĐKXĐ: `1/2 < x < 5`

`log_3(2x-1)*ln(-x+5)=log_(1/27)(2x-1)`

`<=>` `log_3(2x-1)*ln(-x+5)=log_(3^(-3))(2x-1)`

`<=>` `log_3(2x-1)*ln(-x+5)=-1/3log_3(2x-1)`

`<=>` `log_3(2x-1)*[ln(-x+5)+1/3]=0`

TH1: `log_3(2x-1)=0` `<=>` `x=1` (Thỏa mãn)

TH2: `ln(-x+5)+1/3=0`

`<=>` `ln(-x+5)=-1/3`

`<=>` `-x+5=e^(-1/3)`

`<=>` `x = 5 - e^(-1/3)` (Thỏa mãn)

Vậy ................................................