Given exponential functions :
To find which function has initial value of 5, substitute with x = 0 at each function
a)
[tex]\begin{gathered} y=67\cdot(\frac{2}{3})^x \\ \\ x=0\rightarrow y=67\cdot(\frac{2}{3})^0=67\cdot1=67 \end{gathered}[/tex]
b)
[tex]\begin{gathered} y=1.3(6)^x \\ x=0\rightarrow y=1.3\cdot1=1.3 \end{gathered}[/tex]
C)
[tex]\begin{gathered} y=5(0.8)^x \\ \\ x=0\rightarrow y=5(0.8)^0=5\cdot1=5 \end{gathered}[/tex]
d)
[tex]\begin{gathered} y=0.2(5)^x \\ \\ x=0\rightarrow y=0.2\cdot1=0.2 \end{gathered}[/tex]
So, the answer is option c
