# How do you determine if the equation y=90(5)^x represents exponential growth or decay?

Mar 18, 2016

#### Explanation:

If b in ${e}^{b x}$ is > 0, ${e}^{b x}$ grows exponentially..

y = 90${e}^{x \ln 5}$. Here, the factor 90 > 0.

Also, ln 5 =1.609.. > 0

Explanation 2 :
A growth is said to be exponential if the rate of change of growing quantity is proportional to that quantity itself.

i.e. \frac{dy}{dx} \prop y; \qquad \frac{dy}{dx} = \kappa y; \qquad \frac{dy}{y} = \kappa dx,

Integrating this,
$\setminus \ln \left(y\right) = \setminus \kappa x + c , \setminus \quad$where $c$ is an arbitrary constant.
$y \left(x\right) = \setminus \exp \left(\setminus \kappa x + c\right) = \setminus \exp \left(c\right) . \setminus \exp \left(\setminus \kappa x\right) = A \setminus \exp \left(\setminus \kappa x\right)$

Because $A = \setminus \exp \left(c\right)$ and the exponential function is positive for any finite value of $x$, $A$ must be necessarily positive.

If $\setminus \kappa < 0$, it is exponential decay and if $\setminus \kappa > 0$ it is exponential growth.

Exponential growths are characterised by constant doubling times given by ${T}_{2} = \setminus \ln \frac{2}{\setminus} \kappa$

To verify if $y = f \left(x\right)$ represent exponential growth take the first derivative of $y$ with respect to $x$ and see if it is proportional to $y$ itself.
y=90\times5^x; \qquad => \ln(y) = \ln(90)+x\ln(5);
1/y\frac{dy}{dx} = \ln(5); \qquad \frac{dy}{dx} = \ln(5).y; \qquad\frac{dy}{dx} \prop y

Alternatively you can see if $f \left(x\right)$ can be cast into the form $A \setminus \exp \left(\setminus \kappa x\right)$ with $A$ and $\setminus \kappa$ both having positive values.

This has been done in the previous explanation where it is shown that $f \left(x\right) = 90 \setminus \times {5}^{x} = 90 \setminus \exp \left(\setminus \ln \left(5\right) . x\right)$

$A = 90 > 0$; $\setminus \quad$ and $\setminus \quad \setminus \kappa = \setminus \ln \left(5\right) > 0$

So this is exponential growth with a doubling time of ${T}_{2} = \setminus \ln \frac{2}{\setminus} \ln \left(5\right)$.