I need help with a math q involving finding relative growth rate and finding an expression

Q: A bacteria culture grows with constant relative growth rate. The bacteria count was 400 after 2 hrs and 25 600 after 6 hrs.

(a) What is the relative growth rate? (Express answer as a percentage)

(b) What was the initial size of the culture?

(c) Find an expression for the number of bacteria after t hours.



Can someone explain how to solve this question?

a)

means that we are going to k as the following:

A(t) = A0 * e^( k * t ) ----> A(2) = 400
A(2) = A0 * e^( k * 2 )
400 = A0 * e^( 2k )
400 / e^( 2k )

A(t) = A0 * e^( k * t ) ----> A(6) = 25,600
A(6) = A0 * e^( k * 6 )
25,600 = A0 * e^( 6k )
25,600 / e^( 6k ) = A0

A0 = A0
25,600 / e^( 6k ) = 400 / e^( 2k )
25,600 / 400 = e^( 6k ) / e^( 2k )
64 = e^( 4k )
ln( 64 ) = 4k
ln( 64 ) / 4 = k <------- answer

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b)

since we have found the growth rate as k = ln( 64 ) / 4, we are going to use any of the above two equations as the following:

A(t) = A0 * e^( k * t ) ----> A(2) = 400 & k = ln( 64 ) / 4
A(2) = A0 * e^( (ln( 64 ) / 4) * 2 )
400 = A0 * e^( (ln( 64 ) / 2) )
400 = A0 * e^( ( (1/2) * ln( 64 )) )
400 = A0 * e^( ln( 64^(1/2) )) )
400 = A0 * e^( ln( 8) )
400 = A0 * 8
400 / 8 = A0
50 = A0 <------ initial size

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c)

expression:

A(t) = 50 * e^( (ln( 64 ) / 4) * t )

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