You can write ${\left( {\frac{1}{2}} \right)^x}$ as ${2^{ - x}}$ as: ${\left( {\frac{1}{2}} \right)^x} = {({2^{ - 1}})^x} = {2^{ - x}}$
But what about ${\left( {\frac{2}{3}} \right)^x}$? Can it be written so that it has a negative $x$ exponent? I am asking this because for base 'b' of an exponential function where $0<b<1$ is makes sense for me to write it this way as it is intuitive, i.e. ${\left( {\frac{1}{2}} \right)^x} = {2^{ - x}}$, it is intuitive because it allows me to think of the graph of this function as the reflection of the graph of the function $f(x) = {2^x}$ in the y-axis. Is it possible for me to to think of functions such as: $${\left( {\frac{2}{3}} \right)^x},{\left( {\frac{4}{9}} \right)^x},{\left( {\frac{2}{5}} \right)^x},{\left( {\frac{6}{7}} \right)^x},etc$$ In this way?
It comes down to $a^{-1}=\frac{1}{a}$. This gives you $$ 2^{-x} = (2^x)^{-1} = \frac{1}{2^x} = \left(\frac 1 2\right)^x.$$ From this you also get $$ \left(\frac{2}{3}\right)^x = \left(\frac{3}{2}\right)^{-x}. $$