How do you divide #(8.1*10^2)/(9.0*10^-3)#?

2 Answers
Mar 29, 2015

You can deal separately with the numeric and the exponential part. So, first of all, divide #8.1# by #9.0#, obtaining #0.9#. Then, for the exponential part, the exponent will be the sum of the powers at the numerator, minus the powers to the denominator. So, you have #10^{2-(-3)}=10^{2+3}=10^5#.

The final answer is thus
#\frac{8.1 * 10^2}{9.0 * 10^{-3}} = 0.9 * 10^5#

Mar 29, 2015

Divide the coefficients and then the exponents separately before recombining.
#(8.1*10^2)/(9.0*10-3)#

#= (8.1)/(9.0) xx (10^2)/(10^(-3))#

#= 0.9 xx (10^2 * 10^3)#

#= 0.9 xx 10^5#

which would usually be normalized as:

#9.0 xx 10^4#