If b in #e^(bx)# is > 0, #e^(bx)# grows exponentially..

y = 90#e^(xln 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,

#\ln(y) = \kappa x + c, \quad #where #c# is an arbitrary constant.

#y(x) = \exp(\kappa x + c) = \exp(c).\exp(\kappa x) = A\exp(\kappa x)#

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

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

Exponential growths are characterised by constant doubling times given by #T_2=\ln(2)/\kappa#

To verify if #y=f(x)# 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(x)# can be cast into the form #A\exp(\kappax)# with #A# and #\kappa# both having positive values.

This has been done in the previous explanation where it is shown that #f(x)=90\times5^x = 90\exp(\ln(5).x)#

#A = 90 > 0#; #\quad# and #\quad\kappa = \ln(5) > 0#

So this is exponential growth with a doubling time of #T_2=\ln(2)/\ln(5)#.