How do you simplify #2^3xx2^-5#?

2 Answers

Answer:

#2^-2 = 1/4#

Explanation:

We will use the rules of indices.

We know, #a^m xx a^n = a^(m + n)# and #a^-m = 1/a^m#
So, Here,

#2^3 xx 2^-5 = 2^(3 + (-5)) = 2^(3 - 5) = 2^-2 = 1/4#

We can use other laws of indices here.

#a^m/a^n = a^(m-n)#

Using this,

#2^3 xx 2^-5 = 2^3/2^5 = 2^(3 - 5) = 2^-2 = 1/4#

And there are other ways too...

Hope this helps.

Mar 9, 2018

Answer:

Using the same format as the question: #color(blue)(2^(-2))#

Two raise to the power of negative two.

Explanation:

#color(blue)("The shortcut approach:")#

Write as #2^(3-5) = 2^(-2)#

#"====================================="#
#color(blue)("Explaining why this works by using first principles")#

#color(brown)("Before we start this is an important 'concept' (way of thinking)")#

A whole number is what is called a 'rational number'. This means that it can be written in the fractional format of #a/b# where #b# is not 0. The way we do this is for example:

#1,2,3,4,5 ->1/1,2/1,3/1,5/1# and so on. I will use this to emphasise a point.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Answering the question")#

Note that another way of writing #2^(-5)# is #1/2^5#. They both mean the same thing.

Putting this back together we have:

#2^3xx1/2^5#

#(2^3)/1xx1/2^5#

#(2^3xx1)/(1xx2^5) = 2^3/2^5#

#(2xx2xx2)/(2xx2xx2xx2xx2)# we can 'split' this

#ubrace((2xx2xx2)/(2xx2xx2))color(white)("dd")xxcolor(white)("dd")1/(2xx2)#
#color(white)("ddd")darr#

#color(white)("d.dd")ubrace(1color(white)("dddd.d")xxcolor(white)("dd")1/(2xx2))#

#color(white)("dddddddddd")1/2^2#

Another way of writing #1/2^2# is #2^(-2)#

Also note that #1/2^2=1/4=4^(-1)#

So #4^(-1)=2^(-2)#

Hope this helps!