I'm not sure, but I think you mean #(2xx10^4)(3xx10^5)# (Use two x, 'xx' inside to get x instead of #x#
You can change the order of multiplication, so you'll multiply the parts that do not involve #10# to a power separately from the parts that do involve #10# to a power:
#(2xx10^4)(3xx10^5)=2xx10^4xx3xx10^5#
#=2xx3xx10^4xx10^5#
#=(2xx3)xx(10^4xx10^5)#
#=6xx(10^(4+5))#
#=6xx10^9#
Here's another example that skips some steps:
#(3xx10^5)(4xx10^8)#
#=(3xx4)xx(10^5xx10^8)#
#=12xx(10^(5+8))#
#=12xx10^13#
But that is not in scientific notation, so we re-write it:
#12=1.2xx10^1# So
#12xx10^13=1.2xx10^1xx10^13#
#1.2xx(10^(1+13))#
The final answer is:
#1.2xx10^14#
