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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 $1$.

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    $\begingroup$ $a\cdot r^{-1}$ is $\frac{a}{r}$, but without the ${}=1$. $\endgroup$
    – GEdgar
    Jul 20, 2016 at 14:07
  • $\begingroup$ Think about how the exponential function is defined. What is the definition of $r^{-1}$? $\endgroup$
    – Ashvin Swaminathan
    Jul 20, 2016 at 14:07
  • $\begingroup$ The graph is of $y=ar^x$. If the point $(-1,1)$ is on the graph, that means $(-1,1)$ is a solution to that equation, which means we plug in $-1$ for $x$ and $1$ for $y$. This gives $1=ar^{-1}$. Similarly you should get $5=ar^{-2}$ for a second equation. $\endgroup$
    – anon
    Jul 20, 2016 at 14:16

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$r^{-1} = \frac{1}{r}$

That is why.

This is because an exponent is the number of times the same number is multiplied together. So: $$x^4=x\times x\times x\times x$$

If we then divide by $x$, this is the same as multiplying by $x^{-1}$

Because $$\frac{x\times x\times x\times x}{x} = x^4\times x^{-1} = x^{4-1} = x^3$$

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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 $a$.

$1=a\cdot 5\Rightarrow a=\frac15$

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    $\begingroup$ Thanks a ton for your help, @callculus! I really appreciate the time you took to help me with such a simple problem! <3 $\endgroup$
    – Carlos Carlsen
    Jul 20, 2016 at 14:38
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    $\begingroup$ @CarlosCarlsen You´re welcome. It pleases me that the problem is now simple for both of us. $\endgroup$
    – callculus42
    Jul 20, 2016 at 14:43
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You must recall that $r^{-1} = 1/r$ is the reciprocal of $r$, i.e., the unique number such that $rr^{-1} = 1$.

You are given the general form $h(x) = ar^x$. From $1 = h(-1) = ar^{-1}$ we get $a = r$, and thus $h(x) = a^{x+1}$. Then from $5 = h(-2) = a^{-2+1} = a^{-1}$ we get $a = 1/5$.

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  • $\begingroup$ Ok, I guess I am mainly confused with how to get rid of that $-1$ in $a \cdot r ^{-1}$ $\endgroup$
    – Carlos Carlsen
    Jul 20, 2016 at 14:23
  • $\begingroup$ @CarlosCarlsen You can multiply both sides of the equation $1 = ar^{-1}$ by $r$ to obtain $r = 1\cdot r = ar^{-1}\cdot r = a \cdot 1 = a$. $\endgroup$
    – Alex Provost
    Jul 20, 2016 at 14:25

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