Solutions to Exponential and Logarithm Problems

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)

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