Question

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

Answer 1

`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`

Answer 2

`?=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`

Answer 3

`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`