# How do you simplify and write 3^8 xx 3^0 xx 3^1 with positive exponents?

Apr 5, 2018

${3}^{9}$

#### Explanation:

This is the same as ${3}^{8 + 0 + 1} = {3}^{9}$

Note that ${3}^{0} = 1$

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$\textcolor{b l u e}{\text{Why is that so?}}$

Suppose we had ${a}^{3} / {a}^{2}$ then this is the same as ${a}^{3 - 2} = {a}^{1} = a$

$\frac{a \times a \times a}{a \times a} = \frac{a}{a} \times \frac{a}{a} \times a = 1 \times 1 \times a = a$

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Now lets change that a bit.

We know that ${a}^{2} / {a}^{2} = \frac{a \times a}{a \times a} = \frac{a}{a} \times \frac{a}{a} = 1 \times 1 = 1$

but there is another way of writing ${a}^{2} / {a}^{2}$ and that is ${a}^{2 - 2} = {a}^{0}$

So a value that is written as ${a}^{0}$ is 1 no matter what the value of $a$ is.

It is argued that ${0}^{0} = 1$

My professor tolled me this and who am I to argue.

Apr 5, 2018

${3}^{9}$

#### Explanation:

${3}^{8} \times {3}^{0} \times {3}^{1}$

$\therefore = {3}^{\left(8 + 0 + 1\right)}$

$\therefore = {8}^{9}$