Given the relationship between y and x to be
[tex]y=a^x\text{ ------ equation 1}[/tex]Take the logarithm of both sides,
[tex]\begin{gathered} \log y=\log ^{}_{}a^x \\ \Rightarrow\log \text{ y = x }\times\text{ log a ---- equation 2} \end{gathered}[/tex]But when x = 8, log y = 1.603.
Thus, substituting the above values into equation 2, we have
[tex]\begin{gathered} 1.603\text{ = 8 }\times\text{ log a} \\ \text{divide both sides by 8} \\ \log \text{ a= }\frac{1.603}{8} \\ \Rightarrow\log \text{ a =0.2}004 \\ \text{Thus, } \\ a=1.586 \end{gathered}[/tex]From equation 1,
[tex]\begin{gathered} y=a^x \\ \Rightarrow y=1.586^x\text{ ----- equation 3} \end{gathered}[/tex]Thus, when x = 8
[tex]\begin{gathered} y=1.586^x \\ y=1.586^8 \\ \Rightarrow y=40.03 \end{gathered}[/tex]Thus, the value of y will be 40 (to the nearest whole number)
The correct option is D