In this equation A is constant.
I am trying to convert it into an algebraic equation so I can easily solve for "x".
When I graphed it I noticed it looked like something I should be able to match to an easy function but I am unable to.
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1$a^{-1}=\frac{1}{a}$ are you familiar with geometric sums? – Ahmed S. Attaalla May 24 '17 at 22:45
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What do you mean by "convert it into an algebraic equation"? To have an equation, you have to have an "=" sign and a right-hand-side. – MasB May 24 '17 at 22:51
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You can use the formula for the first n terms of a geometric series:
$$\sum_{k=0}^{n-1}cr^k = c\frac{1-r^n}{1-r}$$
Since $a^{-n} = (\frac{1}{a})^n$, we have $r = \frac{1}{a}$ and $c = 1$. So our result is $\displaystyle\frac{1-a^{-x-1}}{1-a^{-1}}$.
orlp
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If $S$ is your sum, then \begin{align} S &= a^0 + a^{-1} + a^{-2} + \cdots + a^{-x} \\ a^{-1} S&= \qquad a^{-1} + a^{-2} + \cdots + a^{-x} + a^{-(x+1)} \end{align} If you subtract these two equations you get \begin{align} S(1-a^{-1}) &= a^0 - a^{-(x+1)} \\ S &= \frac{1 - a^{-(x+1)}}{1-a^{-1}} \end{align}
angryavian
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