# Solve: `log_3 (3^x-1)*log_3(3^(x+1)-3)=6`

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The equation `log_3 (3^x-1)*log_3(3^(x+1)-3)=6` has to be solved for x.

`log_3 (3^x-1)*log_3(3^(x+1)-3)=6`

=> `log_3 (3^x-1)*log_3(3^x*3-3)=6`

=> `log_3 (3^x-1)*log_3(3*(3^x - 1))=6`

=> `log_3 (3^x-1)*(1 + log_3(3^x - 1))=6`

Let `log_ 3(3^x - 1) = y`

=> `y*(1 + y) = 6`

=> `y + y^2 - 6 = 0`

=> `y^2 - 3y + 2y - 6 = 0`

=> `y(y - 3) + 2(y - 3) = 0`

=> `(y + 2)(y - 3) = 0`

=> `y = -2` and `y = 3`

`log_ 3(3^x - 1) = -2`

=> `3^x - 1 = 1/9`

=> `3^x = 10/9`

=> `x = log(10/9)/(log 3)`

`log_ 3(3^x - 1) = 3`

=> `3^x - 1 = 27`

=> `3^x = 28`

=> `x = log 28/log 3`

**The solution of the equation is `log(10/9)/(log 3)` and **`log 28/log 3`

is the number one in the first parentheses and number three in the second separate from those three-pointers on x and x +1??