Finding $b$ such that $e^{5B_t - bt}$ is a martingale I have $X_t = e^{5B_t}$ and 
Where $B_t$ is brownian motion at time $t$.
$M_t = X_t \cdot e^{-bt}$
I need to find a value for $b$ such that $M_t$ is a martingale.
I am encountering difficulty, however.
$$\mathbb{E}[ e^{5B_t}e^{-bt} | \mathcal{F}_s] \space \text{for} \space s\leq t$$
$$= e^{-bt}\mathbb{E}[ e^{5B_t} | \mathcal{F}_s]$$ 
$$= e^{-bt}\mathbb{E}[ e^{5(B_t-B_s)+5B_s} | \mathcal{F}_s]$$ 
$$=exp\{\frac{25(t-s)}{2}+5B_s-bt\}$$
Now since $M_t$ is a martingale if $E[M_t | \mathcal{F}_s] = M_s$
We require that $$exp\{\frac{25(t-s)}{2}+5B_s-bt\} = exp\{5B_s-bs\}$$
Isn't this impossible to solve? Or have I made a mistake?

Let $f(t,x)$ be the function defined by $f(t,x) = e^{5x - bt}$. We observe that $\partial_t f(t,x) = -b f(t,x)$ and $\partial_x^2 f(t,x) = 25 f(t,x)$. For $(X_t)_{t \geq 0}$ to be a local martingale, we require that $\partial_t f(t,x) = -\frac{1}{2} \partial_x^2 f(t,x)$. This tell us that for $(X_t)_{t \geq 0}$ to be a local martingale, we require that $b = \frac{25}{2}$. Let us check whether this value of $b$ also gives us that $(X_t)_{t \geq 0}$ is a martingale. To this end, we need to check that for all $T> 0$, $$\mathbb{E} \left( \int_0^T \left| \partial_x f(t,x) \right|^2 dt \right) < \infty.$$ So observe that \begin{eqnarray*}
\mathbb{E} \left( \int_0^T \left| \partial_x f(t,x) \right|^2 dt \right) &=& 25 \int_0^T \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi t}} \exp \left( - \frac{x^2}{2} \right) \exp \left( 5x - \frac{25}{2} t \right) dx dt \\
&=& \frac{25}{\sqrt{2\pi}} \int_0^T \frac{1}{\sqrt{t}} \exp \left( - \frac{25}{2}t \right) \int_{-\infty}^{\infty} \exp \left( 5x - \frac{1}{2} x^2 \right) dx dt \\
&=& 25 e^{\frac{25}{2}} \int_0^T \frac{1}{\sqrt{t}} \exp \left( - \frac{25}{2} t \right) dt < \infty. 
\end{eqnarray*} Therefore $(X_t)_{t \geq 0}$ defines a martingale. 
