# How many viruses are there after seven days?Scientists culturing a virus find the number of viruses given by the function P(z)=2000+6000 lg(z + a), z is the number of days. If there were originally...

How many viruses are there after seven days?

Scientists culturing a virus find the number of viruses given by the function P(z)=2000+6000 lg(z + a), z is the number of days. If there were originally 2572 viruses,

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P(z) = 2000 + 6000 lg(z+a) where z is number of days and P is the number if viruses.

From the first == z = 0, They already have 2572 viruses

P(0) = 2572 = 2000 + 600 log (a)

==> 572 = 600log a

==> 572 / 600 = log (a)

==> log a =

==> a = 9 (approx.)

Now after 7 days: z= 7

P(7) = 2000 + 600 log (7 + 9)

= 2722.47**Then after 7 days there will ve 2722 bacteria**

The number of live viruses after z number of days is given by the function P(z)=2000+6000 lg(z + a). To start with we had 2572 viruses.

Therefore P(0) = 2572 = 2000 + 6000 lg(0 + a)

=> 572 = 6000 log a

=> log a = 572 / 6000

=> a = 10^( 572 / 6000)

After 7 days the number of viruses is equal to

P(7) = 2000 + 6000* lg ( 7 + 10^(572/6000))

=> 7497

**The number of virus after 7 days is 7497.**

The number of live virus after z days is P(z) and this number is given by:

P(z) = 2000+6000lg(z+a).

The initial number of virus when z = 0, is given by P(0) = 2000 +6000 log (0+a).

P(0) = 2572 given.

Therefore 2000 + 6000 log(0+a) = 2572 .

6000 loga = 2572 - 2000 = 572.

Therefore loga = 572/6000 .

Therefore a = 10^(572/6000) = 1.24547

Therefore the given equation becomes:

P(z) = 2000+6000log(z+1.24547).

Therefore the number of viruses after 7 days= P(7) = 2000+6000log(7+1.24547) = 2000+6000log(8.24547) = 2000+6000(0.916215) = 2000+5497 = 7497

So the number of virus after 7 days = 7497.