Given: 2E2 x 2E-4 These two figures are written in E-notation.
2E2 indicates 2x10^2 in exponential/power-of-ten notation, and 2E-4 does 2x10^-4,so
2E2 x 2E-4=(2x10^2)(2x10^-4) ··· Eq.1
First, multiply two coefficients in front of exponentials, 2 times 2, and then multiply exponential components, 10^2 times 10^-4, and finally multiply the two products obtained.
Use the rules for exponents: a^m x a^n=a^(m+n) where a,m and n are any real
numbers, and a^-m=1/a^m (a is not 0), so Eq.1 can be rewritten:
2E2 x 2E-4=(2 x 2)(10^2 x 10^-4)=4x10^(2-4)=4x10^-2=4/10^2=4/100=0.04
The answer is: 2E2 x 2E-4=4E-2 (=4x10^-2=0.04)
The solutions shown above are all for explanations. You can solve this question just multiplying coefficients, and then exponentials separately, e.g.
2E3 x 6E-5=(2x6)E(3-5)=12E-2=1.2E-1 (=1.2x10^-1=0.12)