# How do you determine if the equation f(x) = 5(1/6)^x represents exponential growth or decay?

Feb 17, 2016

Logic corrected!$\text{ "-> " decay}$

#### Explanation:

Total rewrite!

Consider the case of: $\text{ } x < 0$

We have: $\frac{5}{{\left(\frac{1}{6}\right)}^{x}} = \frac{5 \times {6}^{x}}{{1}^{x}} = 5 \times {6}^{x}$

Thus as the magnitude of negative $x$ increases the value of y also increases.

So as {x: x < 0} increases (moved to the right) it is the reverse of what I previously wrote, so we have decay.

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Consider the case of $\text{ } x = 0$

We have: $\text{ } 5 {\left(\frac{1}{6}\right)}^{0} = 5 \times 1 = 5$

also we have:

$5 {\left(\frac{1}{6}\right)}^{- 0} > 5 {\left(\frac{1}{6}\right)}^{0} > 5 {\left(\frac{1}{6}\right)}^{+ 0}$

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Consider the case of $\text{ } 0 < x$

The denominator of $\frac{1}{6}$ increases as $x$ increases in ${\left(\frac{1}{6}\right)}^{x}$
So ${\left(\frac{1}{6}\right)}^{x}$ becomes smaller and smaller as $x$ increases.

This means that $5 {\left(\frac{1}{6}\right)}^{2}$ decreases as $x$ increases.

So again we have decay

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In conclusion: the equation is in decay