Giúp tôi giải bài tập trên

Đáp án: `S={10^{\frac{1+\sqrt{4log2020+5}}{2}};10^{\frac{1-\sqrt{4log2020+5}}{2}}}`

Giải thích các bước giải:

 `log_{2}x-log(2020x)-1=0` `(1)`

Điều kiện: `x>0`

`(1)<=>log_{2}x-log(2020.x)-1=0`

`<=>log_{2}x-(log2020+logx)-1=0`

`<=>log_{2}x-logx-log_{2020}-1=0`

`<=>(logx)^{2}-2. logx. 1/2+(1/2)^{2}-log2020-1-(1/2)^{2}=0`

`<=>(logx-1/2)^{2}-log2020-5/4=0`

`<=>(logx-1/2)^{2}=log2020+5/4`

`<=>(logx-1/2)^{2}=\frac{4log2020+5}{4}`

`<=>(logx-1/2)^{2}=(\pm\frac{\sqrt{4log2020+5}}{2})^{2}`

`+)TH1:logx-1/2=\frac{\sqrt{4log2020+5}}{2}`

`logx=\frac{1+\sqrt{4log2020+5}}{2}`

`x=10^{\frac{1+\sqrt{4log2020+5}}{2}}(tmđk)`

`+)TH2:logx-1/2=-\frac{\sqrt{4log2020+5}}{2}`

`logx=\frac{1-\sqrt{4log2020+5}}{2}`

`x=10^{\frac{1-\sqrt{4log2020+5}}{2}}(tmđk)`

Vậy `S={10^{\frac{1+\sqrt{4log2020+5}}{2}};10^{\frac{1-\sqrt{4log2020+5}}{2}}}`