How do you evaluate #a^ { 5} \times ? = a ^ { 9}#?

3 Answers
Jul 1, 2017

#a^4#

Explanation:

#x^a*x^b=x^(a+b)#

Let's take a=2;b=3

#x^2*x^3-=(x*x)(x*x*x)-=x*x*x*x*x=x^5-=x^(2+3)#

In this case #?=a^9/a^5=(a*a*a*a*a*a*a*a*a)/(a*a*a*a*a)=(a*a*a*a*cancel(a*a*a*a*a))/(cancel(a*a*a*a*a))=a*a*a*a=a^4#

Jul 1, 2017

#?=a^4#

Explanation:

For this question you need to know the following index law:

#a^x/a^y=a^(x-y)#

Divide both sides by #a^5#

#?*a^5=a^9#

#?*a^5/a^5=a^9/a^5#

#rArr?*cancel(a^5)/cancel(a^5)=a^9/a^5#

Now apply the index law:

#rArr?=a^9/a^5=a^(9-5)=a^4#

Jul 1, 2017

#a^5xxa^4=a^9#

Explanation:

We are given that #a^5# is to be multiplied by something to result in #a^9#.

To find out what the multiplier is , we can divide the two numbers given.

For example: if #5*x=10#,

then #5x=10; so: x=10/5; and: x=2#

where we divided both sides by 5: #x/5# and #10/5#

Here, we will divide #a^9# by #a^5# to find #?#.

We know that since both of these unknown numbers are represented by #a#, our answer will be some multiple of #a#.

We also know that #a^9=a*a*a*a*a*a*a*a*a#

But, these are exponents and we know that dividing exponents means subtract.

So: #a^9/a^5=a^(9-5)=a^4#

And: #a^5xxa^4=a^9#