Question

for any numbers x,y [x=0 in(4) and y = 0 in (5)] and any positive integers m,n, the following holds:x^m · x^n=x^m+nProve number 1

Answer 1

Proved

Explanation:

To prove x^m · x^n=x^m+n, let's assign numbers to x, m and n

let x = 2

m = 3, n = 4

x^m · x^n = 2^3 . 2^4

x^m+n = 2^(3+4)

Solve each of the above seperately and comparew the answer:

$\begin{gathered} x^m\times x^n=2^3\times2^4 \\ =\text{ (2}\times2\times2)\times(2\times2\times2\times2) \\ =\text{ 8}\times16 \\ =\text{ }128 \end{gathered}$$\begin{gathered} x^{m+n}=2^{3+4} \\ =2^7\text{ = 2}\times2\times2\times2\times2\times2\times2 \\ =\text{ 128} \end{gathered}$$\begin{gathered} sincex^m\times x^n\text{ = 128} \\ \text{and x}^{m+n}\text{ = 128} \\ \text{Therefore, }x^m\times x^n\text{ = x}^{m+n} \end{gathered}$

This expression x^m · x^n=x^m+n has been proved to be equal