How do you determine if $y = 30^{- x}$ $y=30^-x$ is an exponential growth or decay?

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Answer 1 · 2016-12-20T01:11:09

Exponential decay

$y = 30^{- x}$ $y=30^-x$

Since the exponent is negative we can deduce that $y$ $y$ decreases for increasing $x$ $x$ . Hence $y$ $y$ represents decay.

However, we can be more rigorous in the analysis.

$ln y = - x ln 30$ $lny=-xln30$

$\frac{1}{y} \frac{d y}{d x} = - ln 30$ $1/y dy/dx = -ln30$

$\frac{d y}{d x} = - ln 30 \cdot 30^{- x}$ $dy/dx = -ln30* 30^-x$

Since $30^{- x} > 0 \forall x$ $30^-x >0 forall x$

$\frac{d y}{d x} < 0$ $dy/dx <0$ over the domain of $y$ $y$

Since $\frac{d y}{d x}$ $dy/dx$ gives the slope of $y$ $y$ at any point $x$ $x$ in its domain, $y$ $y$ is always decreasing for increasing $x$ $x$

Hence $y$ $y$ represents exponential decay.

How do you determine if y=30^-xy=30−xy=30^-x is an exponential growth or decay?