solve e^x = 1 + 6e^−x

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`e^x = 1+6e^(-x)`

`e^x=1+6/(e^x)`

Multiplying by `e^x` ,

`e^(2x)=e^x+6`

`e^(2x)-e^x-6 = 0`

Let `e^x = t` , then,

`t^2-t-6 = 0`

We can solve this for t.

`(t+2)(t-3)=0`

Therefore, `t=-2` or `t =3` .

When `t =-2` ,

`e^x = -2`

This is not true, since `e^x` is not less than zero. Then `t!= -2` and `e^x!=-2`

When `t=3,`

`e^x = 3`

Taking log,

`x =ln 3`

**Therefore the solution is `x =ln 3` .**

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