Solutions to Exponential and Logarithm Real Life 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:
\[
5000 e^{10a} = 10000
\]
Divide both sides of the equation by 5000 and simplify:
\[
e^{10a} = \dfrac{10000}{5000}
\]
\[
e^{10a} = 2
\]
Using \(\ln\), the above exponential form can be written as:
\[
10a = \ln 2
\]
Solve the above equation for \( a \) and approximate the answer to 3 decimal places:
\[
a = \dfrac{\ln 2}{10} \approx 0.069
\]
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 (\( P_1 = P_2 \)) when \( t = 20 \).
\[
10000 e^{20k} = 20000 e^{20 \times 0.01}
\]
Divide both sides of the equation by \( 20000 e^{20k} \) and simplify:
\[
\dfrac{10000}{20000} = \dfrac{e^{0.2}}{e^{20k}}
\]
\[
\dfrac{1}{2} = e^{0.2 - 20k}
\]
The above exponential form may be converted into logarithmic form as follows:
\[
0.2 - 20k = \ln\left(\dfrac{1}{2}\right)
\]
Solve for \( k \):
\[
k = \dfrac{0.2 - \ln\left(\dfrac{1}{2}\right)}{20}
\]
\[
k \approx 0.045 \quad \text{(to three decimal places).}
\]
Solution to Problem 3:
If \( I = 10^{-8} \ \text{watts/cm}^2 \), then
\[
D = 10 \log\left(\dfrac{I}{10^{-16}}\right)
\]
\[
= 10 \log\left(\dfrac{10^{-8}}{10^{-16}}\right)
\]
\[
= 10 \log(10^{8}) = 80 \ \text{decibels}.
\]
Solution to Problem 4:
Using the decibel formulas from problem 3, we can write
\[
60 = 10 \log\!\left(\dfrac{I_1}{10^{-16}}\right)
\]
and
\[
80 = 10 \log\!\left(\dfrac{I_2}{10^{-16}}\right).
\]
Subtract the above equations side by side as follows
\[
60 - 80 = 10 \log\!\left(\dfrac{I_1}{10^{-16}}\right) - 10 \log\!\left(\dfrac{I_2}{10^{-16}}\right).
\]
The above may be written
\[
20 = 10 \left[ \log\!\left(\dfrac{I_2}{10^{-16}}\right) - \log\!\left(\dfrac{I_1}{10^{-16}}\right) \right].
\]
Divide both sides by 10 and use the logarithm property
\(\log a - \log b = \log \left(\dfrac{a}{b}\right)\)
to rewrite the above equation as follows
\[
2 = \log\!\left(\dfrac{I_2}{I_1}\right).
\]
The above gives
\[
\dfrac{I_2}{I_1} = 10^2 = 100.
\]
Solution to Problem 5:
If the number of people infected is \( N = 2000 \). We use this information to write the following equation:
\[
2000 = \dfrac{15000}{1 + 100 e^{-0.5 t}}
\]
To find the number of days, we need to solve the above equation for \( t \).
Rewrite the above equation as follows:
\[
1 + 100 e^{-0.5 t} = \dfrac{15000}{2000}
\]
Which can be written as:
\[
e^{-0.5 t} = \dfrac{\dfrac{15000}{2000} - 1}{100}
\]
We now use the natural logarithm \(\ln\) to rewrite the above equation:
\[
-0.5 t = \ln \left( \dfrac{\dfrac{15000}{2000} - 1}{100} \right)
\]
Now solve for \( t \):
\[
t = \dfrac{\ln \left( \dfrac{\dfrac{15000}{2000} - 1}{100} \right)}{-0.5} \approx 5.467
\]
(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
\[
A(t) = A_{0} e^{-kt},
\]
which gives an initial amount equal to \(A_{0}\). Using the above definition of half life, we can write
\[
A_{0} e^{-kT} = \dfrac{A_{0}}{2}.
\]
Divide both sides by \(A_{0}\) to obtain
\[
e^{-kT} = \dfrac{1}{2}.
\]
Use \(\ln\) to change the exponential form into logarithmic form:
\[
-kT = \ln\!\left(\dfrac{1}{2}\right).
\]
Simplify \(\ln\!\left(\dfrac{1}{2}\right)\) and solve for \(T\):
\[
T = \dfrac{\ln 2}{k}.
\]
Solution to Problem 7:
\[
y = \dfrac{a}{1 + b e^{-kt}}
\]
Multiply both sides by \( 1 + b e^{-kt} \) and divide both sides by \( y \):
\[
1 + b e^{-kt} = \dfrac{a}{y}
\]
Subtract \(1\) from both sides:
\[
b e^{-kt} = \dfrac{a}{y} - 1
\]
Divide both sides by \(b\):
\[
e^{-kt} = \dfrac{\dfrac{a}{y} - 1}{b}
\]
Rewrite the exponential form into logarithmic form:
\[
-kt = \ln\!\left(\dfrac{\dfrac{a}{y} - 1}{b}\right)
\]
Finally, solve for \(t\):
\[
t = -\dfrac{1}{k} \ln\!\left(\dfrac{\dfrac{a}{y} - 1}{b}\right)
\]
Solution to Problem 8:
Use the given information to write the equation
\[
V_{o} e^{-at} = 0.5 V_{o}
\]
Divide both sides by \( V_{o} \):
\[
e^{-at} = \dfrac{1}{2}
\]
Use \(\ln\) to change exponential form into logarithmic form:
\[
- at = \ln\!\left(\dfrac{1}{2} \right) = - \ln 2
\]
Solve for \( t \):
\[
t = \dfrac{\ln 2}{a}
\]
Solution to Problem 9:
We substitute the ratio in decibels and \( P_i \) in the given formula
\[
10 = 10 \log \left( \dfrac{10}{P_i} \right)
\]
Divide both sides by 10
\[
1 = \log \left( \dfrac{10}{P_i} \right)
\]
Rewrite the logarithmic form into exponential form
\[
\dfrac{10}{P_i} = 10^1 = 10
\]
Solve for \( P_i \)
\[
P_i = 1 \, \text{mW}
\]
Solution to Problem 10:
Substitute \( t = \dfrac{L}{R} \) in the given formula for the current \( i \):
\[
i = \dfrac{E}{R} \left( 1 - e^{-\dfrac{R \cdot L}{L \cdot R}} \right)
\]
Simplify:
\[
i = \dfrac{E}{R} \left( 1 - e^{-1} \right) \approx 0.632 \, \dfrac{E}{R}
\]
More math problems with detailed solutions in this site.