If b in e^(bx) is > 0, e^(bx) grows exponentially..
y = 90e^(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).
