Logarithms Part II
Application of logarithms to real world problems
Example 3
The value of a car decreases so that after a period of n year, the value of a car is $\ 50000e^{-0.1n} $.
i) Find the value of a newly bought car
ii) Find the value of the car after 10 years
iii) The car is scraped when its value drops to 10000 dollars. Determine when the car will be scraped.
Let v represent the value of a car after n years
Note that v and n are variables
i) When a car is newly bought, n = 0
$ \begin{aligned}
v &= 50000e^{-0.1n} \\
v &= 50000e^{-0.1*0} \\
v &= 50000 \quad \textbf{note that e to the power of zero is 1}\\
\end{aligned}$
The value of a newly bought car is 50000 dollars
ii) Value of a car after 10 years, n = 10
$ \begin{aligned}
v &= 50000e^{-0.1n} \\
v &= 50000e^{-0.1*10} \\
v &= 18400 \quad \text{ (to 3 s.f.)}\\
\end{aligned}$
The value of the car after 10 years is 18400 dollars
iii)
Technique used: Taking log on both sides
$ \begin{aligned}
10000 &= 50000e^{-0.1n} \\
\frac{10000}{50000} &= e^{-0.1n} \\
\ln {0.2} &= \ln {e^{-0.1n}\\
-1.609 &= -0.1n \\
n &= 16.1 \\
\end{aligned}$
The car has to be scraped after 16.1 years
Example 4
The value of a house, V dollars after t years is determined by $\ V = V_0 e^{kt} $, where $\ V_0 $ is the original value of the house and k is a constant.
i) Determine k if the value of the house doubles after 5 years
ii) After how many years would the price of the house be 10 times its original value
Note that V and t are variables. k is a constant. In other words, no matter what the value of V and t is, the value of k does not change
i)
Technique used: Taking log on both sides
After 5 years, the value of the house is $\ 2V_0 $
$ \begin{aligned}
2V_0 &= V_0 e^{5k} \\
\frac{2V_0}{V_0} &= e^{5k} \\
2 &= e^{5k} \\
\ln {2} &= \ln {e^{5k}\\
0.6931 &= 5k \\
k &= 0.139 \quad \text{(to 3 s.f.)} \\
\end{aligned}$
ii)
Technique used: Taking log on both sides
When price of house is 10 times its original value, $\ V = 10V_0 $
$ \begin{aligned}
10V_0 &= V_0 e^{0.139t} \\
\frac{10V_0}{V_0} &= e^{0.139t} \\
10 &= e^{0.139t} \\
\ln {10} &= \ln {e^{0.139t}\\
2.3026 &= 0.139t \\
t &= 16.6 \quad \text{(to 3 s.f.)} \\
\end{aligned}$
After 16.6 years, the price of the house will be 10 times its original value
As you can see, the "Taking log on both sides" technique is commonly used.
Logarithms Part III will show you how to solve equations involving logarithms.
Example 3
The value of a car decreases so that after a period of n year, the value of a car is $\ 50000e^{-0.1n} $.
i) Find the value of a newly bought car
ii) Find the value of the car after 10 years
iii) The car is scraped when its value drops to 10000 dollars. Determine when the car will be scraped.
Let v represent the value of a car after n years
Note that v and n are variables
i) When a car is newly bought, n = 0
$ \begin{aligned}
v &= 50000e^{-0.1n} \\
v &= 50000e^{-0.1*0} \\
v &= 50000 \quad \textbf{note that e to the power of zero is 1}\\
\end{aligned}$
The value of a newly bought car is 50000 dollars
ii) Value of a car after 10 years, n = 10
$ \begin{aligned}
v &= 50000e^{-0.1n} \\
v &= 50000e^{-0.1*10} \\
v &= 18400 \quad \text{ (to 3 s.f.)}\\
\end{aligned}$
The value of the car after 10 years is 18400 dollars
iii)
Technique used: Taking log on both sides
$ \begin{aligned}
10000 &= 50000e^{-0.1n} \\
\frac{10000}{50000} &= e^{-0.1n} \\
\ln {0.2} &= \ln {e^{-0.1n}\\
-1.609 &= -0.1n \\
n &= 16.1 \\
\end{aligned}$
The car has to be scraped after 16.1 years
Example 4
The value of a house, V dollars after t years is determined by $\ V = V_0 e^{kt} $, where $\ V_0 $ is the original value of the house and k is a constant.
i) Determine k if the value of the house doubles after 5 years
ii) After how many years would the price of the house be 10 times its original value
Note that V and t are variables. k is a constant. In other words, no matter what the value of V and t is, the value of k does not change
i)
Technique used: Taking log on both sides
After 5 years, the value of the house is $\ 2V_0 $
$ \begin{aligned}
2V_0 &= V_0 e^{5k} \\
\frac{2V_0}{V_0} &= e^{5k} \\
2 &= e^{5k} \\
\ln {2} &= \ln {e^{5k}\\
0.6931 &= 5k \\
k &= 0.139 \quad \text{(to 3 s.f.)} \\
\end{aligned}$
ii)
Technique used: Taking log on both sides
When price of house is 10 times its original value, $\ V = 10V_0 $
$ \begin{aligned}
10V_0 &= V_0 e^{0.139t} \\
\frac{10V_0}{V_0} &= e^{0.139t} \\
10 &= e^{0.139t} \\
\ln {10} &= \ln {e^{0.139t}\\
2.3026 &= 0.139t \\
t &= 16.6 \quad \text{(to 3 s.f.)} \\
\end{aligned}$
After 16.6 years, the price of the house will be 10 times its original value
As you can see, the "Taking log on both sides" technique is commonly used.
Logarithms Part III will show you how to solve equations involving logarithms.
Comments
Post a Comment