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

Jul 1, 2017

${a}^{4}$

Explanation:

${x}^{a} \cdot {x}^{b} = {x}^{a + b}$

Let's take a=2;b=3

${x}^{2} \cdot {x}^{3} \equiv \left(x \cdot x\right) \left(x \cdot x \cdot x\right) \equiv x \cdot x \cdot x \cdot x \cdot x = {x}^{5} \equiv {x}^{2 + 3}$

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

${a}^{5} \times {a}^{4} = {a}^{9}$

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 $5 \cdot x = 10$,

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

where we divided both sides by 5: $\frac{x}{5}$ and $\frac{10}{5}$

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 ${a}^{9} = a \cdot a \cdot a \cdot a \cdot a \cdot a \cdot a \cdot a \cdot a$

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

So: ${a}^{9} / {a}^{5} = {a}^{9 - 5} = {a}^{4}$

And: ${a}^{5} \times {a}^{4} = {a}^{9}$