Assuming that x does not equal zero, translate the following according to exponent law:
5x/5x
\(x\ {\in \mathbb R }\)\(\) \ 0
\(\frac{5x}{5x}=5^1\times x^1 \times 5^{-1}\times x^{-1}=5^{1-1}\times x^{1-1}\)
\(=5^0\times x^0 =1\times 1=1\)
\(\frac{5x}{5x}=1\) 
!
