Why is $a\cdot r^{-1}=1$ equivalent to $\frac {a}{r} = 1$?
I am trying to write exponential functions from graphs; two points were given: $(-1,1)$ & $(-2,5)$.
I am trying to find an equation for $a$, so I used the point $(-1,1)$ to do so.
I get: $h(-1) = a\cdot r^{-1} = 1$. Then this is where I get stuck -
Don't get me wrong, I have tried to solve this on my own, but I don't see the process by which I can solve this and the course I am doing does not explain it well :/
I guess this is a very specific question relating to exponents because I searched on Google 'how to get rid of an exponent' (yes, I know... I don't know what I am talking about, but it worked!) and the results I recieved showed equations using a logarithim, so this must relate to the powers $-1$ and $1$.
A general form of a exponential function can be defined as $y(x)=a\cdot r^x$ with $r>0$.
Now you can insert the given x/y values.
$1= a\cdot r^{-1}$
$5= a\cdot r^{-2}$
Now divide the first equation by the second equation.
$\frac{1}{5}=\large{\frac{a\cdot r^{-1}}{a\cdot r^{-2}}}$
The parameter $a$ can be cancelled out. Note that $r^{-1}=\frac1r$ and $r^{-2}=\frac1{r^2}$. Therefore $\large{\frac{ r^{-1}}{r^{-2}}=\frac{ r^{2}}{r^{1}}=r}$.
Thus $r=\frac{1}{5}$
And with the help of the first equation we can calculate the value of $a$.
$1=a\cdot 5\Rightarrow a=\frac15$