$4^{Brownian(t)}$ martingale proof Let $B(t)$ a Brownian motion.
I like to prove that $4^{B(t)}$ = martingale
I rewrote the expression into an exponential form (like $\exp(\ln(4) B)$), but then I don't know how to proceed.
 A: Recall that
$$\mathbb{E}e^{c B_t} = e^{\frac{1}{2} c^2 t} \tag{1}$$
as $B_t$ is Gaussian with mean $0$ and variance $t$. In particular, we see that
$$M_t := 4^{B_t} = \exp \bigg( B_t \cdot \log 4 \bigg)$$
is not a martingale since
$$\mathbb{E}M_t \stackrel{(1)}{=} \exp \left( \frac{1}{2} (\log 4)^2 \cdot t \right)$$
is not constant. In fact, by the stationarity and independence of the increments, we have
$$\begin{align*} \mathbb{E}(M_t \mid \mathcal{F}_s) &= M_s \cdot \mathbb{E}(e^{(B_t-B_s) \log 4} \mid \mathcal{F}_s) \\ &= M_s \cdot \mathbb{E}e^{B_{t-s} \log 4} \\ &\stackrel{(1)}{=} M_s \cdot e^{\frac{1}{2}(\log 4)^2 (t-s)}. \end{align*}$$
Note that this calculation shows that
$$\tilde{M}_t := \exp \bigg( B_t \cdot \log 4- \frac{1}{2} (\log 4)^2 \cdot t \bigg)$$
is a martingale.
A: Let $M(t)$ your expression, let $F_s = \sigma\{M(u), u\le s\}$
Start with the beginning: $\forall t>0\ \ E [M(t)]<\infty$
Then, prove that $\forall t>s\ge0 \ \ E[M(t)|F_s] = M_s$.
The general method is to prove that for all 
measurable function $A$ and times $0\le s_1\le\ldots\le s_d = s<t$
$$
E[A(M(s_1),\ldots, M(s_d))M(t)] = E[A(M(s_1),\ldots, M(s_d))M(s)]
$$
Here, with $M(t) = \frac{4^{B(t)}}{E[4^{B(t)}]}$, you just chek that this is well defined:
$$
E[4^{B(t)}] = \int  \exp\left(\ln 4\sqrt{t}y-\frac{y^2}{2}\right) \frac{dy}{\sqrt{2\pi}}<\infty
$$because the square dominates in the exponential.
Then:
$$
E\left[A(M(s_1),\ldots, M(s_d))\frac{4^{B(t)}}{E[4^{B(t)}]}\right] 
= \frac{1}{E[4^{B(t)}]}
E\left[A(M(s_1),\ldots, M(s_d)) 4^{B(s)} 4^{B(t) - B(s)}
\right] 
\\=
\frac{1}{E[4^{B(t)}]}
E\left[A(M(s_1),\ldots, M(s_d)) 4^{B(s)} \right]
E\left[4^{B(t) - B(s)}
\right] 
$$
because $B(t) - B(s)$ and $F_s$ are independant.
For the same reason you have $$
E\left[4^{B(t)}\right]  = E\left[4^{B(t)-B(s)}\right] 
E\left[4^{B(s)}\right]
$$
and you eventually get
$$
E\left[A(M(s_1),\ldots, M(s_d))\frac{4^{B(t)}}{E[4^{B(t)}]}\right] =
E\left[A(M(s_1),\ldots, M(s_d))\frac{4^{B(s)}}{E[4^{B(s)}]}\right] 
$$
