\[100 \times 10^x=10^{2y}\]
\[\frac{10 \times 10^x}{10^y}=100\]
We can write the first of the as
\[10^2 \times 10^x=10^{2y} \rightarrow 2+x=2y\]
We can write the second of these as
\[\frac{10^1 \times 10^x}{10^y}=10^2 \rightarrow 1+x-y=2\]
We now have the equations
\[2+x=2y\]
(1)\[1+x-y=2\]
(2)Thje first of these minus the second gives
\[(2+x)-(1+x-y)=2y-2 \rightarrow 1+y=2y-2 \rightarrow 1+2=2y-y \rightarrow y=3\]
.Substiture
\[y=3\]
into (1) then \[x=2y-2=4\]