[tex]\boxed{f(x)=3(3)^x}[/tex]
A landscaper put 9 lilypads into a new pond. We know that the number of lilypads triples each month over a period of time. That is:
[tex]\text{For month 1} \rightarrow \text{Number of lilypads}=9 \\ \\ \text{For month 2} \rightarrow \text{Number of lilypads}=9\times 3=27 \\ \\ \text{For month 3} \rightarrow \text{Number of lilypads}=27\times 3=81 \\ \\ \text{For month 4} \rightarrow \text{Number of lilypads}=81\times 3=243[/tex]
And so one. Therefore, this follows a geometric sequence at which the constant ratio is 3. This is also modeled as an exponential function, so we can write it as follows:
[tex]f(x)=ar^{x-1} \\ \\ \\ a:\text{Number of lilypads for the first month} \\ \\ x: \text{Number of month} \\ \\ f(x):The number of lilypads \\ \\ \\ So: \\ \\ a=9 \\ \\ r=3 \\ \\ f(x)=9(3)^{x-1} \\ \\ \\ By \ property: \\ \\ a^{n-1}=a^n/a \\ \\ \\ Then: \\ \\ f(x)=9(\frac{3^x}{3}) \\ \\ f(x)=\frac{9}{3}(3^x) \\ \\ \boxed{f(x)=3(3)^x}[/tex]