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$$ \frac{100}{9} = \frac{1 - \frac{1}{(1+x)^{12}}}{x} $$

I tried and tried, but can't seem to get $x$ into a form to isolate it or use a quadratic formula or imaginary numbers or something.

I need to know how to do this for an exam, so I have to do it by hand using a regular calculator. What steps are to be taken to get the answer?

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  • $\begingroup$ Oh, thank you. It is perfectly correct still. $\endgroup$ – Johnny May 22 '14 at 4:07
  • $\begingroup$ That won't go anywhere near nice... $\endgroup$ – chubakueno May 22 '14 at 4:13
  • $\begingroup$ How to solve for x? Cry. $\endgroup$ – Vincent May 22 '14 at 5:04
  • $\begingroup$ Have you considered the possibility that the 12 might be a typo? Considering the context, I suspect it's supposed to be a 2. $\endgroup$ – hasnohat May 22 '14 at 5:14
  • $\begingroup$ No, it's supposed to be '12', not '2'. $\endgroup$ – Johnny May 22 '14 at 5:22
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Rough Plot: $$1-\dfrac{1}{(1+x)^{12}},\dfrac{100x}{9}$$

This shows you that there are two solutions. One, after discontinuity and before root near $x=-1$ and the other positive solutions very near $0$. Note that $x=0$ is not a solution as it is not in domain of original expression.

For the small positive root, Use binomial approximation : $(1+x)^{-12}\approx1-12x+78x^2$

It gives $0.011396$ very near to the root. You can check it in your calculator.

Next we have to find the root just between $-1$ and $-2$. You can use your calculator to do this. Find value of $$1-\dfrac{1}{(1+x)^{12}},\dfrac{100x}{9}$$ for values $\{-2,-1.9...\}$

You will note that the sign changes near$-1.7$. Then you can approximate it more decimal places. While calculating, you can simplify your calculations using $1+x=t$

$$100(t-1)/9=1-1/t^{12}$$

$$100t^{13}=109t^{12}-9$$

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