# Is the equation V(T)=100(1.08)^T an exponential growth or decay?

Check the BASE of your exponential; $1.08$ is bigger than $1$ so for increasing $T$ it gets bigger.
$T = 0 \to {1.08}^{T} = {1.08}^{0} = 1$
$T = 1 \to {1.08}^{T} = {1.08}^{1} = 1.08$ so the value of your function is increasing.