Question

Given an exponential function for compounding interest, A(x)=P(1.02)^x, what is the rate of change?

Answer 1

Answer:Its 2% guys

Explanation:

Answer 2

`\bf \qquad \textit{Amount for Exponential Growth}\\\\ A=I(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ I=\textit{initial amount}\\ r=rate\to r\%\to (r)/(100)\\ t=\textit{elapsed time}\\ \end{cases}\\\\ -------------------------------\\\\ \stackrel{A}{A(x)}=\stackrel{I}{P}(1.02)^{\stackrel{t}{x}}\implies A=P(1+\stackrel{r}{0.02})^x\implies \cfrac{r}{100}=0.02 \\\\\\ r=100\cdot 0.02\implies r=\stackrel{\%}{2}`