Other applications of exp and log

Exponential and logarithmic growth: Future value

Other applications of exp and log

We will now look at some other applications of exponential functions and logarithms.

First we will look at a frequently used rule of thumb in economic calculations.

Rule of thumb for the growth rate of a product

If two variables each show a relatively small percentage growth, the product of these two variables will grow with a rate approximately equal to the sum of the growth rates of the individual variables.

Or: For small percentages the growth rate of the product of two variables is approximately equal to the sum of the growth rates of the individual variables.

In order to understand why this rule of thumb works, we assume that #i# and #j# are small growth rates. The corresponding growth factors are #1+i# and #1+j#, respectively, so the growth factor of the product of two variables, say #A# with growth rate #i# and #B# with growth rate #j#, is equal to

\[(1+i)\cdot(1+j)=1+i+j+i\cdot j\]

The fact that #i# and #j# are small implies that #i\cdot j# is very small. If, for instance #i\approx 10^{-2}# and #j\approx 10^{-2}#, then #i\cdot j\approx 10^{-4}#, so if we round numbers to 2 decimals, we can forget about #i\cdot j#. Thus, for sufficiently small values of #i# and #j#, we can take the growth factor of #A\cdot B# to be #1+i+j#, and so the grow rate to be #i+j#, as stated.

Since, the growth percentages corresponding to #i#, #j# and #i+j# are each obtained from the growth rates by multiplication by #100#, the same rule of thumb holds for growth percentages instead of growth rates. After all,

\[100\cdot i+100\cdot j = 100\cdot(i+j)\]

Suppose that the price #P# of a product grows by #7\%# and the sales #Q# (the number of products sold) grows by #5\%#.

By what percentage does the revenue #P \cdot Q# grow according to the rule of thumb?
By what percentage does the revenue grow by a precise calculation?

Round your answers to #2# decimals.

According to the rule of thumb: #12# #\%#
According to precise calculation: #12.35# #\%#

The rule of thumb says that if #P# and #Q# each show a fairly small percentage of growth, then the turnover #P\cdot Q# grows at a percentage equal to the sum of the growth percentages of the individual variables. So the turnover grows by approximately #7+5=12\%#.

For a precise calculation, we argue as follows:
\[P_{\text{new}}=P_{\text{old}} \cdot 1.07 \text{ and } Q_{\text{new}}=Q_{\text{old}} \cdot 1.05 \]
so:
\[(P \cdot Q)_{\text{new}}=(P\cdot Q)_{\text{old}} \cdot 1.07 \cdot 1.05 =(P \cdot Q)_{\text{old}} \cdot 1.1235\]
So the turnover #P \cdot Q# increased by precisely #12.35\%#.

The difference between the calculation with the rule of thumb and the precise answer is #0.35\%#. This difference is negligible in a reasoning like: "if the price of the butter increases by #7\%#, and the marketing by #5\%#, then the turnover grows by #12\%#".

This difference is so small because both growth rates are small. If the growth rates are higher, using the rule of thumb is very risky. For example, suppose #P# grows by #70\%# and #Q# grows by #50\%#, then according to the rule of thumb, the turnover grows by #120\%#, but according to a precise calculation by #155\%#. This is a substantial deviation.

New example

II.

As we have seen, the calculation of growth in a savings account on the basis of compound interest uses an exponential growth model. We can apply the same model to the study of population growth. In the examples below, we will pose three questions related to population growth and answer these by means of the theory of exponential growth and logarithms.

Suppose the population growth in a given country is #0.5\%# per year.

What is the growth model for the population at a unit interval of a year?

The growth model is: #B(n)=B_0 \cdot (1.005)^n#, where #B(n)# is the population after #n# years and #B_0# is the population at the beginning.

After all, the fact that the population growth is #0.5\%# per annum, means that the growth rate at which the population grows each year, is equal to #i_{\text{year}}=\frac{0.5}{100}=0.005#. Consequently, the annual growth factor is equal to: #g_{\text{year}}=0.005+1=1.005#. In conclusion, we can use the growth model:
\[B(n)=B_0 \cdot (1.005)^n\] where #B(n)# is the population after #n# years and #B_0# the population at the beginning.

New example

III.

In order to describe the annual growth of a process, we can use the average annual decrease/increase. This is the percentage of decrease/increase per year computed from the decrease/increase in terms of a percentage over an extended period under the assumption that the process follows an exponential growth model. Here is an example.

Suppose that the decrease in unemployment follows an exponential growth model and that unemployment decreased by #27\%# over #4# years.

What percentage is the average annual decrease?

Round your answer to an integer.

Average annual decrease: #8##\%#

The growth rate for #4# years is #\frac{-27}{100}=-0.27#, so the growth factor for #4# years equals #-0.27+1=0.7300#.

Since #\frac{1}{4}# of #4# years is the same as a year, the theory Converting growth factors shows that the growth factor per year is obtained by raising the growth factor for #4# years to the power #\frac{1}{4}#. Therefore, the growth factor per year is #(0.7300)^{\frac{1}{4}}=0.9243#.

Consequently, the growth rate per annum is equal to #0.9243-1=-0.0757#, and the corresponding percentage is #-0.0757 \cdot 100=-7.57#. Thus the average annual decrease in unemployment is approximately #8\%#.

New example

IV.

Finally, we will look at the so-called 70-rule, which is a rule of thumb for the doubling time in an exponential growth model. We will also formulate a similar rule for the tripling time.

70 rule

The amount of capital in a bank account growing with compound interest #p\%#, where #p\lt 10#, has an approximate doubling time of #70# divided by the interest percentage per year:

\[n_{\text{double}}\approx \frac{70}{p}\]

To prove the rule we use the formula #S(n)=S_0 \cdot (1+\frac{p}{100})^n# for the final value after #n# years with growth rate #\frac{p}{100}#. Doubling the capital in #n# years means #S(n)=2 \cdot S_0#.

Substituting the value for #S(n)# in the formula gives the equation: \[2 \cdot S_0=S_0 \cdot (1+\frac{p}{100})^n\] Dividing both sides by #S_0#, we find the equation \[2 = (1+\frac{p}{100})^n\] We solve this equation for #n#, because we are interested in the time it takes for the capital at the bank to double.

We will use the following approximation rule for small #x#, which we will not prove here: \[\log_{10}(1+x)\approx \frac{x}{2.30258509}\]

The solution is as follows:

\[\begin{array}{rcl}
(1+\frac{p}{100})^n&=&2\\
&& \phantom{xxxxx}\color{blue}{\text{the equation to be solved}}\\
\log_{10}(1+\frac{p}{100})^n&=&\log_{10}(2)\\
&& \phantom{xxxxx}\color{blue}{\text{logarithm applied to both sides}}\\
n \cdot \log_{10}(1+\frac{p}{100})&=&\log_{10}(2)\\
&& \phantom{xxxxx}\color{blue}{\text{rule }\log_g(x^n)=n \cdot \log_g(x)}\\
n&=& \frac{\log_{10}(2)}{\log_{10}(1+\frac{p}{100})}\\
&& \phantom{xxxxx}\color{blue}{\text{divided by }\log_{10}(1+\frac{p}{100})}\\
n&\approx& \frac{\log_{10}(2)}{\frac{p}{100 \cdot 2.30258509}}\\
&& \phantom{xxxxx}\color{blue}{\text{approximation rule for small }p \text{: } \log_{10}(1+\frac{p}{100})=\frac{p}{100 \cdot 2.30258509}}\\
n&\approx&\frac{69.31471806}{p}\\
&& \phantom{xxxxx}\color{blue}{\text{simplified}}\\
\end{array}\]
Rounding #69.31471806# to #70#, we find that for small #p#, \[n_{double}\approx\frac{70}{p}\]

For tripling we also have a similar rule.

110 rule

The amount of capital in a bank account growing with compound interest, has an approximate tripling time of #110# divided by the annual interest percentage #p#, where #p\lt 10#:

\[n_{\text{triple}}\approx \frac{110}{p}\]

The proof is the same as the doubling rule. Again we use the formula #S(n)=S_0 \cdot (1+\frac{p}{100})^n#, in which the growth rate is #\frac{p}{100}#. We use that #S(0)# will be tripled when #S(n)=3 \cdot S_0#. This leads to the equation: \[3 \cdot S_0=S_0 \cdot (1+\frac{p}{100})^n\] Dividing both sides by #S_0# we find the equation \[3 = (1+\frac{p}{100})^n\] In solving this equation for #n#, we use again (without proof) the approximation rule for small #x# \[\log_{10}(1+x)\approx \frac{x}{2.30258509}\]

\[\begin{array}{rcl}
(1+\dfrac{p}{100})^n&=&3\\
&& \phantom{xxxxx}\color{blue}{\text{the equation to be solved}}\\
\log_{10}(1+\dfrac{p}{100})^n&=&\log_{10}(3)\\
&& \phantom{xxxxx}\color{blue}{\text{logarithm applied to both sides}}\\
n \cdot \log_{10}(1+\dfrac{p}{100})&=&\log_{10}(3)\\
&& \phantom{xxxxx}\color{blue}{\text{rule }\log_g(x^n)=n \cdot \log_g(x)}\\
n&=& \dfrac{\log_{10}(3)}{\log_{10}(1+\frac{p}{100})}\\
&& \phantom{xxxxx}\color{blue}{\text{divided by }\log_{10}(1+\frac{p}{100})}\\
n&\approx& \dfrac{\log_{10}(3)}{\frac{p}{100 \cdot 2.30258509}}\\
&& \phantom{xxxxx}\color{blue}{\text{approximation rule for small }p \text{: }}\\&& \phantom{xxxxx}\color{blue}{\log_{10}(1+\frac{p}{100})=\frac{p}{100 \cdot 2.30258509}}\\
n&\approx&\dfrac{109.8612293}{p}\\
&& \phantom{xxxxx}\color{blue}{\text{calculated}}\\
\end{array}\]
Rounding #109.8612293# to #110#, we find that for small #p#, \[n_{triple}\approx\frac{110}{p}\]

The following example shows how this works in practice.

Someone deposits #\euro \, 1200# in a savings account at #7\%# compound annual interest.
a) What is the doubling time according to the 70 rule?
b) What is the tripling time according to the 110 rule?
c) After how many years will the capital be multiplied by 16?

Round all your answers to whole years.

Doubling time: #10# years
Tripling time: #16# years
Time to grow sixteenfold: #40# years

According to the 70 rule, the doubling time is approximately: \[n_{\text{double}}=\frac{70}{7}=10\]
So after #10# years, the capital is about #\euro \, 2400#.

For estimating the tripling time, we can use the 110 rule: \[n_{\text{triple}}=\frac{110}{7}=16\]
So after #16# years, the capital is about #\euro \, 3600#.

After #4 \cdot 10=40# years the amount has become the sixteenfold of the originial capital, so after #40# years, the capital is approximately equal to #\euro \, 19200#.

For comparison: the exact amount saved after #10# years will be equal to #\euro \, 2360.58#, after #16# years it will be #\euro \, 3474.77#, and after #40# years it will be #\euro \, 17969.35#.

New example