PermalinkSubmitted by mzjp2 on Thu, 12/01/2016 - 01:47
What sort of form do you want to get this into? Presumably you want to simplify it rather than write out all it's decimal digits? If so, remember the fact that $a^m \cdot a^k = a^{m+k}$ for all $m,k$ and a given $a$. This can be "shown" (note it's not an actual proof, it only holds for integer $m$ and $k$) by writing out the terms like so:
$$a^m \cdot a^k = \underbrace{a\cdot a\cdots a}_{m \, \text{times}} \cdot \underbrace{a \cdot a \cdots a}_{k \, \text{times}} = \underbrace{a \cdot a \cdot a \cdots a}_{(m+k) \, \text{times}} = a^{m+k}$$
Can you spot what $(a,m,k) = (?,?,?)$ you have in your problem?
What sort of form do you want
What sort of form do you want to get this into? Presumably you want to simplify it rather than write out all it's decimal digits? If so, remember the fact that $a^m \cdot a^k = a^{m+k}$ for all $m,k$ and a given $a$. This can be "shown" (note it's not an actual proof, it only holds for integer $m$ and $k$) by writing out the terms like so:
$$a^m \cdot a^k = \underbrace{a\cdot a\cdots a}_{m \, \text{times}} \cdot \underbrace{a \cdot a \cdots a}_{k \, \text{times}} = \underbrace{a \cdot a \cdot a \cdots a}_{(m+k) \, \text{times}} = a^{m+k}$$
Can you spot what $(a,m,k) = (?,?,?)$ you have in your problem?