# How do you simplify (7^4)^3?

Jan 7, 2017

${7}^{12}$

#### Explanation:

Using the $\textcolor{b l u e}{\text{law of exponents}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{a}^{m} \times {a}^{n} = {a}^{m + n}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
which can be extended to the product of more than 2 terms.

$\Rightarrow {\left({7}^{4}\right)}^{3} = {7}^{4} \times {7}^{4} \times {7}^{4}$

$= {7}^{4 + 4 + 4} = {7}^{12}$

Note that the exponent 12, is also obtained by $\textcolor{b l u e}{\text{multiplying}}$ 3 and 4 together. This leads us to a further law of exponents.

$\text{That is } \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{\left({a}^{m}\right)}^{n} = {a}^{m n}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\Rightarrow {\left({7}^{4}\right)}^{3} = {7}^{4 \times 3} = {7}^{12}$

Jan 7, 2017

${\left({7}^{4}\right)}^{3} = {7}^{12}$

#### Explanation:

Note that if $a$ is any number and $m , n$ are positive integers then:

${\left({a}^{m}\right)}^{n} = {\overbrace{{a}^{m} \times {a}^{m} \times \ldots \times {a}^{m}}}^{\text{n times}}$

color(white)((a^m)^n)=overbrace(overbrace(axxaxx...xxa)^"m times"xxoverbrace(axxaxx...xxa)^"m times"xx...xxoverbrace(axxaxx...xxa)^"m times")^"n times"

$\textcolor{w h i t e}{{\left({a}^{m}\right)}^{n}} = {\overbrace{a \times a \times \ldots \times a}}^{\text{mn times}}$

$\textcolor{w h i t e}{{\left({a}^{m}\right)}^{n}} = {a}^{m n}$

So in our example:

${\left({7}^{4}\right)}^{3} = {7}^{4 \cdot 3} = {7}^{12}$