The number of bugs doubles every 4 days. If the bug population starts with 6 bugs, then the equation, A(t) =6(2)^0.25 can represent the number of bugs. Find the value for A(8).

To find the value of \( A(8) \), we need to clarify the equation provided. It appears there is a small confusion in the equation you posted. The correct form for the exponential growth of the bug population that doubles every 4 days should be:

\[ A(t) = A_0 \cdot 2^{t/T} \]

where

• \( A_0 \) is the initial population,
• \( T \) is the period of doubling (4 days in this case),
• \( t \) is the time in days.

Given that the initial population \( A_0 \) is 6, the equation becomes:

\[ A(t) = 6 \cdot 2^{t/4} \]

Now, to find \( A(8) \):

\[ A(8) = 6 \cdot 2^{8/4} \]

Calculating the exponent:

\[ 8/4 = 2 \]

So we have:

\[ A(8) = 6 \cdot 2^2 \]

Calculating \( 2^2 \):

\[ 2^2 = 4 \]

Now, substituting back:

\[ A(8) = 6 \cdot 4 = 24 \]

Thus, the value for \( A(8) \) is \( \boxed{24} \).