Solve for $x$ in $e^x$

Question

I can't get this to work out right and can't find anything that gives me a rule for how to distribute the $0.5$ here. Any help would be much appreciated.

\begin{align*} 0.5 &=\frac{\exp(-3.6 + 1.8x)}{1 + \exp(-3.6 + 1.8x)}\\ &\\ 0.5(1 + \exp(-3.6 + 1.8x)) &= \exp(-3.6 + 1.8x) \end{align*}

Answer 1

You have:

$\dfrac{1}{2}(1 + e^{-3.6 + 1.8x}) = e^{-3.6 + 1.8x}$, so

$e^{-3.6 + 1.8x} = 1$, and taking logs of both sides, we get

$$-3.6 + 1.8 x = 0$$

I think you can manage the rest.

Answer 2

Initially define $y=\exp(1.8x)$ Then your first equation becomes $0.5=\frac {\exp(-3.6)y}{1+\exp(-3.6)y}$ Solve for $y$, then note that $x = \frac 1{1.8}\log y$

Answer 3

OK, so how about seeing that if we write $u=-3.6 + 1.8x$ then we have

\begin{equation} \frac{1}{2}\left(1+e^{u}\right) = e^{u} \end{equation} Now, let's mutiply out the brackets \begin{equation} \frac{1}{2} + \frac{1}{2}e^{u} = e^{u} \end{equation}

so (upon subtracting $\frac{1}{2}e^{u}$ from both sides); \begin{equation} \frac{1}{2} = \frac{1}{2}e^{u} \end{equation} or \begin{equation} 1 = e^{u} \end{equation} Do you think you can solve it now? Then substitute back in for $x$?

Answer 4

Suppose that $y = \text{exp}(-3.6+1.8x)$. Then your original equation reduces to:

$$0.5 = \frac{y}{1+y}$$

Thus, we have:

$$0.5 (1+y) = y$$

which when simplified yields:

$$y=1$$.

Therefore, we have to solve for:

$$\text{exp}(-3.6+1.8x)=1$$

Now, we know that $e^0=1$. You should be able to find $x$ now.