Question

How do you solve `1-2e^(2x)=-19`?

Answer 1

`x = ln \sqrt{10}`

Explanation:

`1 - 2 e^{2x } = -19`

`-2 e^{2x} = -19 -1 = -20`

`e^{2x} = -20/(-2) = 10`

`ln e^{2x} = ln 10`

`2x = ln 10`

`x = {ln 10}/2 = ln \sqrt{10}`

Check:

`1 - 2 e^{2x }`

`= 1 - 2 e^{2 (ln sqrt{10} ) }`

`= 1 - 2 e^{ln 10}`

`= 1 - 2(10)`

`= -19 quad sqrt`

Answer 2

the value is `~~1.151`

Explanation:

given `1-2e^(2x)=-19rArr-2e^(2x)=-20rArre^(2x)=10`
in general we have `e^m=krArr log_ek=m`
which means we have `log_e10=2x` and `log_e10~~2.302`
we have `2x=2.302rArrx~~1.151`

Answer 3

`x = (ln10)/2`
`~~1.1512925465`

Explanation:

Subtract 1 on both sides.
`-2e^(2x) = -20`
Divide by -2.
`e^(2x) = 10`
Taking the logarithm of both sides, we have:
`ln(e^(2x)) = ln10`
Using the power rule of logarithms,
`2xln(e) = ln 10`

`lne = 1` So, we have:

`2x = ln 10`
`x = (ln10)/2`