Solve for t: `xe^(2t) + x =e^(2t) - 1`  

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The equation `xe^(2t) + x =e^(2t) - 1` has to be solved for t.

`xe^(2t) + x =e^(2t) - 1`

=> `x*e^(2t) - e^(2t) = -x - 1`

=> `e^(2t)(x - 1) = (-x - 1)`

=> `e^(2t) = (1 + x)/(1 - x)`

take the natural logarithm of both...

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The equation `xe^(2t) + x =e^(2t) - 1` has to be solved for t.

`xe^(2t) + x =e^(2t) - 1`

=> `x*e^(2t) - e^(2t) = -x - 1`

=> `e^(2t)(x - 1) = (-x - 1)`

=> `e^(2t) = (1 + x)/(1 - x)`

take the natural logarithm of both the sides

`ln e^(2t) = ln((1 + x)/(1 - x))`

=> `2t*ln e = ln((1 + x)/(1 - x))`

=> `2t = ln((1 + x)/(1 - x))`

=> `t = ln((1 + x)/(1 - x))/2`

The solution for t is `t = ln((1 + x)/(1 - x))/2`

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