Algebra to show Induction for $\lceil {x\over 2} \rceil \mapsto \lceil{(x+1)\over 2}\rceil $ $5^{\lceil \frac x 2 \rceil}$ (Which Operation?) x = $5^{\lceil \frac x2\ + 1 ) \rceil}$
$x$ can be any value that can help equate this for induction. 
I'm confused on how the algebra would work to get from $5^{\lceil \frac x 2 \rceil}$ to change it to $(x+1)\over 2$ in the ceiling function itself. 
 A: The wording of the question is still unclear and the notation inconsistent, but I'll take a chance that you are really asking about $5^{\lceil(x+1)/2\rceil}$ and not $5^{\lceil x+(1/2)\rceil}$ on the right-hand side of the equation.
Although you cannot prove a general fact just by looking at a few examples,
sometimes it helps to look at examples anyway, especially if you do not
know exactly what fact you are trying to prove. A few examples can give a hint
about a general pattern.
The ceiling function $\left\lceil \frac x2 \right\rceil$ increases by $1$ whenever $x$ increases by $2$, that is,
$$\left\lceil \frac 12 \right\rceil = 1, \quad
\left\lceil \frac 22 \right\rceil = 1, \quad
\left\lceil \frac 32 \right\rceil = 2, \quad
\left\lceil \frac 42 \right\rceil = 2, \quad
\left\lceil \frac 52 \right\rceil = 3,$$
and so forth.  So a few examples of the known parts of your equation
(the $5^n$ pieces on the left-hand and right-hand sides)
with $x=1,2,3,4,5,6$ gives this:
\begin{align}
5^{\lceil 1/2 \rceil} &= 5   & 5^{\lceil(1+1)/2\rceil} &= 5 \\
5^{\lceil 2/2 \rceil} &= 5   & 5^{\lceil(2+1)/2\rceil} &= 25 \\
5^{\lceil 3/2 \rceil} &= 25  & 5^{\lceil(3+1)/2\rceil} &= 25 \\
5^{\lceil 4/2 \rceil} &= 25  & 5^{\lceil(4+1)/2\rceil} &= 125 \\
5^{\lceil 5/2 \rceil} &= 125 & 5^{\lceil(5+1)/2\rceil} &= 125 \\
5^{\lceil 6/2 \rceil} &= 125 & 5^{\lceil(6+1)/2\rceil} &= 625
\end{align}
You may notice a pattern. Actually you should notice two patterns, one
for the cases where $x$ is even and a different pattern for the cases
where $x$ is odd.
It is perfectly OK for a fact to have separate statements for when $x$ is even
and when $x$ is odd, or for an inductive proof to have two different cases
for the inductive step, one for $n$ even and one for $n$ odd.
