Question

Can someone help me solve for x? (Exponential Equation)

`50/(1 +e^-x) = 4`

Answer 1

`x=-ln 46`

Explanation:

Given:

`50/(1+e^(-x)) = 4`

Multiply both sides by `1+e^(-x)` to get:

`50 = 4+4e^(-x)`

Subtract `4` from both sides to get:

`46 = e^(-x)`

Take natural logs of both sides to get:

`ln 46 = -x`

Multiply both sides by `-1` to get:

`-ln 46 = x`

That is:

`x = -ln 46`

(which is the same as `ln (1/46)`)

Answer 2

`x=-2.44" to 2 dec. places"`

Explanation:

`color(blue)"cross-multiply"` the equation.

`rArr4(1+e^-x)=50`

`"divide both sides by 4"`

`(cancel(4)(1+e^-x))/cancel(4)=50/4`

`rArr1+e^-x=12.5`

`"subtract 1 from both sides"`

`cancel(1)cancel(-1)+e^-x=12.5-1`

`rArre^-x=11.5`

`"using "color(blue)"law of logarithms"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(logx^nhArrnlogx)color(white)(2/2)|)))`

`"take ln (natural log ) of both sides"`

`lne^-x=ln11.5`

`rArr-xcancel(lne)^1=ln11.5`

`rArrx=-ln11.5`

`rArrx~~-2.44" to 2 dec. places"`