__Solution to Problem 1:__

If t = 0 corresponds to 1980 then 1990 corresponds to t = 10. We can say that when t = 10 the population was 10000 and this gives the following equation:

5000 e^{ 10 a} = 10000

Divide both sides of the equation by 5000 and simplify

e^{ 10 a} = 10000 / 5000

e^{ 10 a} = 2

Using ln, the above exponential form can be written as

10 a = ln 2

Solve the above equation for a and approximate the answer to 3 decimal places.

a = ln 2 / 10 = 0.069 (approximated to 3 decimal places).

__Solution to Problem 2:__

If t = 0 corresponds to the year 2000 then t = 20 corresponds to the year 2020. We need to find k so that the two populations are equal (P1 = P2) when t = 20.

10000 e^{ 20 k} = 20000 e^{ 20 * 0.01}

Divide both sides of the equation by 2000 e^{ 20 k} and simplify

10000 / 20000 = e^{ 20 * 0.01} / e^{ 20 k}

1 / 2 = e^{ 0.2 - 20 k}

The above exponential form may be converted in logarithmic form as follows

0.2 - 20 k = ln (1/2)

Solve for k

k = [0.2 - ln (1/2)] / 20 = 0.045

(approximated to 3 decimal places).

__Solution to Problem 3:__

If I = 10^{ -8} watts/cm^{ 2} then

D = 10 log( I / 10^{ -16} )

= 10 log( 10^{ -8} / 10^{ -16} )

= 10 log (10^{ 8}) = 80 decibels.

__Solution to Problem 4:__

Using the decibel formulas from problem 3, we can write

60 = 10 log( I1 / 10^{ -16} )

and

80 = 10 log( I2 / 10^{ -16} )

Subtract the above equations side by side as follows

60 - 80 = 10 log( I1 / 10^{ -16} ) - 10 log( I2 / 10^{ -16} )

The above may be written

20 = 10 [ log( I2 / 10^{ -16} ) - log( I1 / 10^{ -16} ) ]

Divide both sides by 10 and use the logarithm property: log x - log x = log (x / y) to rewrite the above equation as follows

2 = log [I2 / I1]

The above gives

I2 / I1 = 10^{ 2} = 100.

__Solution to Problem 5:__

If the number of people infected is N = 2000. We use this information to write the following equation:

2000 = 15000 / [ 1 + 100 e^{ -0.5 t} ]

To find the number of days we need to solve the above equation for t.

Rewrite the above equation as follows

[ 1 + 100 e^{ -0.5 t} ] = 15000 / 2000

Which can be written as follows

e^{ -0.5 t} = [ 15000 / 2000 - 1 ] / 100

We now use the natural logarithm ln to rewrite the above equation as follows

-0.5 t = ln [ ( 15000 / 2000 - 1 ) / 100 ]

We now solve for t

t = ln [ ( 15000 / 2000 - 1 ) / 100 ] / -0.5 = 5.467 (approximated to 3 decimal places).

__Solution to Problem 6:__

The half life (T) of a radioactive material, is the period of time after which the amount of this material decays to half its initial amount. The initial amount is found by setting t = 0 in the formula A(t) = A_{ o} e ^{ -k t}. Which gives an initial amount equal to A_{ o}. Using the above definition, we can write

A_{ o} e ^{ -k T} = A_{ o} / 2.

Divide both sides by A_{ o} to obtain

e ^{ -k T} = 1/2

Use ln to change the exponential form into logarithmic form

-k T = ln(1/2)

Simplify ln (1/2) and solve for T.

T = ln 2 / k

__Solution to Problem 7:__

Given

y = a / [1 + b e ^{ -k t}]

Multiply both sides by 1 + b e ^{ -k t} and divide both sides by y to obtain

1 + b e ^{ -k t} = a / y

Subtract 1 to both sides

b e ^{ -k t} = a / y - 1

Divide both sides by b

e ^{ -k t} = [ a / y - 1 ] / b

Rewrite the exponential form into logarithmic form

-k t = ln ([ a / y - 1 ] / b)

Solve for t.

t = -(1/k) ln ([ a / y - 1 ] / b)

__Solution to Problem 8:__

Use the given information to write the equation

V_{ o} e ^{ -a t} = 50% V_{ o}

Divide both sides by V_{ o}.

e ^{ -a t} = 1/2

Use ln to change exponential form into logarithmic form

-a t = ln(1/2)

Solve for t

t = ln 2 / a

__Solution to Problem 9:__

We substitute the ratio in decibels and Pi in the given formula

10 = 10 Log (10 / Pi)

Divide both sides by 10.

1 = Log (10 / Pi)

Rewrite the logarithmic form into exponential form

10 / Pi = 10^{ 1} = 10

Solve for Pi.

Pi = 1 mw

__Solution to Problem 10:__

Substitute t by L/R in the given formula for the current i.

i = (E/R)[1 - e^{ -RL/LR}]

Simplify.

i = (E/R)[1 - e^{ -1}] = 0.632 (E/R)

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