I'm not sure, but I think you mean (2xx10^4)(3xx10^5)(2×104)(3×105) (Use two x, 'xx' inside to get x instead of xx
You can change the order of multiplication, so you'll multiply the parts that do not involve 1010 to a power separately from the parts that do involve 1010 to a power:
(2xx10^4)(3xx10^5)=2xx10^4xx3xx10^5(2×104)(3×105)=2×104×3×105
=2xx3xx10^4xx10^5=2×3×104×105
=(2xx3)xx(10^4xx10^5)=(2×3)×(104×105)
=6xx(10^(4+5))=6×(104+5)
=6xx10^9=6×109
Here's another example that skips some steps:
(3xx10^5)(4xx10^8)(3×105)(4×108)
=(3xx4)xx(10^5xx10^8)=(3×4)×(105×108)
=12xx(10^(5+8))=12×(105+8)
=12xx10^13=12×1013
But that is not in scientific notation, so we re-write it:
12=1.2xx10^112=1.2×101 So
12xx10^13=1.2xx10^1xx10^1312×1013=1.2×101×1013
1.2xx(10^(1+13))1.2×(101+13)
The final answer is:
1.2xx10^141.2×1014
