How do I solve this exponential equation problem?

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The exponential growth model is:

#color(red)(A(t) = A_0 e^(kt)#, where

#color(red)(A(t)# represents the amount of bacteria after the amount of time #color(red)(t)#

#color(red)(A_0# represents the amount of bacteria at the beginning of the exponential growth.

#color(red)(k# is a constant value, it varies from problem to problem.

We are given the exponential equation #color(blue)(A(t) = 750e^(0.0425t)#

Hence,

#A_0 = 750#

#k=0.0425#

Now we are ready to answer the questions:

1. What is the initial number of bacteria ?

   Answer: 750
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   2. What is the number of bacteria after 20 minutes ?

   Answer: #A(20)=750e^(0.0425*20)#
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   using the calculator,

#A(20) ~~ 1754.73514#

#A(20) ~~ 1755# ( Rounded to the nearest whole number)

Hence, the number of bacteria after 20 minutes is 1755.
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3. How long will it take for the number of bacteria to double ?

Find the mount of time (in minutes) it takes for the number of bacteria to be 1500.

#A(t) = 750e^(0.0425*t)#

For this question, we know that #A(t) = 1500#

Hence, we must solve for #t# using the exponential equation:

#1500=750e^(0.0425*t)#

To find the value of #t#, divide both sides of the equation by #750#

#1500/750=(750e^(0.0425*t))/750#

#cancel (1500)^color(red)(2)/cancel (750)=(cancel (750)e^(0.0425*t))/cancel(750#

#2=e^(0.0425*t)#

To simplify, take the natural log of both sides of the equation:

#ln 2 = ln e^(0.0425*t)#

Move the exponent to the front and simplify.

#ln 2 =0.0425*t* (ln e)#

Note that #color(blue)(ln e=1#

Hence, we get

#ln 2 =0.0425*t* 1#

#ln 2 =0.0425*t#

To solve for #t#, divide both sides by #0.0425#

We get,

#(ln 2)/0.0425 =(0.0425*t)/0.0425#

#(ln 2)/0.0425 =(cancel 0.0425*t)/cancel 0.0425#

Swap sides:

#t=(ln 2/0.0425)#

#t~~ 16.30934542# (using a calculator)

#t~~ 16.309# (accurate to 3 decimal places)

Hence, it takes 16.309 minutes approximately for the bacteria to double.

Hope you find the solution useful.