Entering first order ode special form ID 1 solver
\[\begin {aligned} x^{\prime }&=b \,{\mathrm e}^{x}\\ x \left (0\right ) &= 1\\ \end {aligned}\]
Writing the ode as
\begin{align*} x^{\prime } &= b \,{\mathrm e}^{x}\tag {1} \end{align*}
And using the substitution \(u={\mathrm e}^{-x}\) then
\begin{align*} u' &= -x^{\prime } {\mathrm e}^{-x} \end{align*}
The above shows that
\begin{align*} x^{\prime } &= -u^{\prime }\left (t \right ) {\mathrm e}^{x}\\ &= -\frac {u^{\prime }\left (t \right )}{u} \end{align*}
Substituting this in (1) gives
\begin{align*} -\frac {u^{\prime }\left (t \right )}{u}&=\frac {b}{u} \end{align*}
The above simplifies to
\begin{align*} u^{\prime }\left (t \right )&=-b\tag {2} \end{align*}
Now ode (2) is solved for \(u \left (t \right )\).
Because the ODE has the form \(u^{\prime }\left (t \right )=f(t)\), the solution requires only integration. Therefore
\begin{align*} du &= \left (-b\right ) \, dt\\ u &= \int { \left (-b\right ) \, dt}\\ &= -t b +c_1 \end{align*}
Substituting the solution found for \(u \left (t \right )\) in \(u={\mathrm e}^{-x}\) gives
\begin{align*} u \left (t \right ) = -t b +c_1 \end{align*}
Applying the initial condition \(x \left (0\right ) = 1\), the solution becomes
\begin{align*}
{\mathrm e}^{-x} &= -t b +{\mathrm e}^{-1} \\
\end{align*}
Solving for \(x\) from \({\mathrm e}^{-x} = -t b +{\mathrm e}^{-1}\) gives
\begin{align*}
x &= -\ln \left (-\left (t b \,{\mathrm e}-1\right ) {\mathrm e}^{-1}\right ) \\
\end{align*}
Summary of solutions found
\[
x = -\ln \left (-\left (t b \,{\mathrm e}-1\right ) {\mathrm e}^{-1}\right )
\]