We are given the following function
[tex](\frac{1}{3})^{2-3t}[/tex]
Let us re-write this function in the following form.
[tex]f(t)=ka^t[/tex]
Where k is a constant.
Step 1:
Split the powers using the multiplication rule of exponents.
[tex]a^{x+y}=a^x\cdot a^y[/tex]
Applying the above rule, the function becomes
[tex](\frac{1}{3})^{2-3t}=(\frac{1}{3})^2\cdot(\frac{1}{3})^{-3t}[/tex]
Further simplifying, the function becomes
[tex](\frac{1}{3})^2\cdot(\frac{1}{3})^{-3t}=\frac{1}{9}\cdot(\frac{1}{3})^{-3t}[/tex]
Step 2:
Apply the power rule of exponents
[tex]a^{xy}=(a^x)^y[/tex]
So, the function becomes
[tex]\frac{1}{9}\cdot(\frac{1}{3})^{-3t}=\frac{1}{9}\cdot((\frac{1}{3})^{-3})^t[/tex]
Further simplifying the function becomes
[tex]\frac{1}{9}\cdot((\frac{1}{3})^{-3})^t=\frac{1}{9}\cdot(27^{})^t[/tex]
Therefore, the function is
[tex]f(t)=\frac{1}{9}\cdot27^t[/tex]
Where k = 1/9 and a = 27
