I have the quaternion $q=a+bi+cj+dk$ and wan't to compute $e^{q}$.
Since $i,j,k$ all behave similar one might assume the following formula works, but why doesn't it?
$$e^{q}=e^{a}\cdot(\cos{b}+i\sin{b})\cdot(\cos{c}+j\sin{c})\cdot(\cos{d}+k\sin{d})$$
Edit: Of course this could be applied for any base $x$ using natural logarithms. (Thanks to Jonas Meyer for pointing this out)
Start with $\;x=e=\exp(1)\,$ and note $\;I:=\dfrac{bi+cj+dk}{\rho}$ with $\;\rho:=\sqrt{b^2+c^2+d^2}\;$ then $\;I^2=-1\;$ and you may write : $$\tag{1}e^{a+\rho I}=e^{\,a}e^{\,\rho I}=e^a\left(\cos(\rho)+I\sin(\rho)\right)$$ The substitution of $\,I\,$ and $\,\rho\,$ giving you the final result.
Concerning the general power of $x$ you may use $\;\displaystyle x^{a+\rho I}=e^{\,\ln(x)(a+\rho I)}$ to conclude.
If $\,x\,$ is a quaternion you may have to define $\;\displaystyle x^q:=e^{\ln(x)\,q}\;$ (for example) first. $$-$$ Since this doesn't answer your question let's add that $(1)$ applies because $\,a\,$ and $\,u:=\rho I\,$ commute (this implies also that a similar method works for generalizations of quaternions like octonions)
so that : $$\tag{2}e^{\,a+u}=e^{\,a}\cdot e^{\,u}$$ Should $\,a\,$ and $\,u\,$ not commute then the Maclaurin expansion of the exponential at the left would for example contain the term $\;\displaystyle\frac {(a+u)^2}2=\frac{a^2+au+ua+u^2}2\,$ which differs from the corresponding terms $\;\displaystyle\frac {a^2}2+au+\frac {u^2}2=\frac{a^2+2au+u^2}2\,$ at the right and $(2)$ would not be true in general.
As indicated by Qiaochu Yuan since $i$, $j$ and $k$ don't commute you can't simply rewrite : $$e^{\,a+bi+cj+dk}=e^a\cdot e^{bi}\cdot e^{cj}\cdot e^{dk}$$ as you did to obtain a solution.
The same way $\,\ln(u\cdot v)=\ln(u)+\ln(v)\,$ is not true in general for quaternions since else you would have : $\,\ln(u\cdot v)=\ln(u)+\ln(v)=\ln(v)+\ln(u)=\ln(v\cdot u)\;$ even for $\;u\cdot v\neq v\cdot u$.