These are detailed solutions to the exponential and logarithm problems.

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)