### 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.

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