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I've been having trouble with these equations. The three equations have no relation to one and other. I've been trying to solve for $x$:

$$y = 1.15^{x+2}$$

$$y = \frac{1}{1.15^{x+2}}$$

$$y = 1 - \frac{1}{1.15^{x+2}}$$

Thank you.

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closed as off-topic by Bookend, Peter Woolfitt, Daniel W. Farlow, ASB, Sujaan Kunalan Mar 22 '15 at 6:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

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Is this supposed to be a simultaneous set of equations? That is, are $x$ and $y$ supposed to satisfy all three equations at once, or are you asking about three different problems? – Zev Chonoles Jan 3 '13 at 5:29
Oops, sorry for not clarifying that. No, they are separate equations with no relationship. – user55003 Jan 3 '13 at 5:29
For the first one use log, for the second one use log, for the third one use log. – Ram Jan 3 '13 at 5:34
up vote 4 down vote accepted

For the first, take the $\log$ of both sides, where $\log$ means logarithm to your favourite base, perhaps $e$, perhaps $10$, perhaps $2$. We get $$\log y=(x+2)\log 1.15.$$ Thus $$x+2=\frac{\log y}{\log 1.15},$$ and now solving for $x$ is straightforward.

For the second, take the reciprocal of both sides. We get $$\frac{1}{y}=1.15^{x+2}.$$ Now use the procedure of the first problem. You may want to use the fact that $\log(1/y)=-\log y$.

Or else you can rewrite as $y=1.15^{-(x+2)}$, and take the logarithm of both sides.

For the third problem, first rewrite as $1-y=\frac{1}{1.15^{x+2}}$.

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Thank you! Your help is appreciated! – user55003 Jan 3 '13 at 5:41

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