Elementary Mathematics
Compound interest formulas Exponential and logarithmic functions
This is an article from my home page: www.olewitthansen.dk
Ole Witt-Hansen 2011
Contents
Chapter 1. Compound interest ...... 1 1. Percentage calculations...... 1 2. The compound interest formula ...... 2 3. The average rate of growth ...... 4 Chapter 2. Savings annuities and debts annuities ...... 5 1. Annuities ...... 5 1.1 Formula for the sum of a geometric series...... 5 1.2 Savings annuity...... 6 1.3 Debts annuities...... 6 Chapter 3. Exponential functions...... 8 1. The concept of a power of a number...... 8 1.1 Calculation with powers...... 8 2. Extension of powers to negative exponents...... 9 3. Extension of powers to rational number exponents...... 10 Chapter 4. Logarithmic functions...... 14 1. Logarithmic functions...... 14 Chapter 5. Eksponential growth and power growth...... 17 1. Exponential growth...... 17 1.1 Solution of exponential equations...... 17 2. Doubling constant and half-live constant...... 18 3. Logaritmic scale...... 19 4. Power functions...... 21
Percentage and interest 1
Chapter 1. Compound interest
1. Percentage calculations The well known percent symbol “%” has been invented to signify 1/100. If we want find 3.5% of 450, then it usually done in the following way (in the grammar school)
1% of 450 is 450/100 = 4.50. 3.5% of 450 = 3.54.50 =15.75. But since 3.5% = 0.035, so that 3.5% of 450 can be directly calculated as 4500.035 = 15.75
From now on we shall always use the short method to calculate a percentages, also for that reason that it is awkward to have units and special symbols in mathematical equations.
Often the letter p is used as the symbol for a certain percentage. p/100 is then denoted r. Thus we have the connection: r = p% = p/100. r is called the interest rate or the growth rate.
When we shall calculate the interest R from the capital K, and the interest rate is r, then we have according to the example above:
(1.1) R = Kr
This formula is of course equally valid, even if it is does not capital capital, but common calculation with percentages. If we want to find p% = r from a quantity k, and the result is i (interest), then we have an equivalent formula:
(1.2) i = kr
However the formula is easier to remember and read, if we keep the letters K, R and r, even if it is not about money so the concepts capital, Interest, and interest rate is to be understood in a broader sense. Using this (mathematical) formula, we may for example answer a question like: How many percent make 27 up of 179. Inserting R = 27 and K = 179 in the formula (1.1) and solving for r. we find:
R 27 R Kr r 0.1508 15.08% K 179
We shall now derive the very important formula, which expresses how much a capital K (or any other quantity having percentage growth) has grown to, when it has increased by r = p% .
If the accumulated capital is K1, then K1 is equal to K plus the interest R = Kr.
(1.3) K1 = K + R = K + Kr = K(1+r), so: K1 = K(1 + r)
Percentage and interest 2
Notice that the formula is equally valid, if we have a percentage decrease instead of a growth. It just means that the interest rate r is negative. That a quantity decreases with 15% is the same as a growth of -15%.
Eample:
1. A department store advertises that the price is reduced by 20 % for a certain product. The product now costs 477.- What was the price before the reduction? We apply the formula (1.5) with K1 = 457, r = - 20% = - 0.2 to determine K.
K = 457/(1-0.2) = 571.25.
2. An enterprise had one year an income of 257.000,- and expenses were 301.000. The following year the income grew with 5%, while the expenses increased by 3%. How many percent has the deficit grown/decreased? If we set the deficit for the two years to u and u1, we have:
u = 257.000 - 301.000 = - 44.000 and u1 = 257.0001.05 - 301.0001.03 = - 40.180
We then apply (1.3): 40.180 = 44.000(1+r) (1+r) = 40180/44000 = 0.9132 then r = 0.9132-1 = -0.0868. Which means that the deficit has been reduced by 8.68 %. 2. The compound interest formula When a capital accumulates interest in a banking institution, the interest is attributable, with a fixed interval called the term. The period between two attributable of interest may be one year, half a year or monthly or whatever has been agreed. But when you talk about the interest rate of a deposit or a loan, the interest rate is always indicated as the yearly interest rate, even if the term is shorter.
If a bank makes a loan to 5% p.a. (pro annum = yearly), and the interest rates are due each half - year, the semi-annually interest rate is 2.5%. As we shall show, however, the effective interest rate will be slightly larger than 5 %. If the capital K has been repaid with the interest rate r in n terms, one might think that the accumulated interest was nKr (n times the interest in one term). This is, however, not the case, since the capital has grown after each term, and therefore the amount due to collect interest has grown. You will also receive interest from the interest.
We shall then establish a formula for Kn , the accumulated capital after n terms when the start capital is K and the interest rate is r. We know already from (1.3) that we can find the capital after one term by multiplying by (1+r). From this follows:
K1 = K(1+r) (The capital after one term) 2 K2 = K1(1+r) = K(1+r) (The capital after two terms) 3 K3 = K2(1+r) = K(1+r) (The capital after three terms) ...... n Kn = Kn-1(1+r) = K(1+r) (The capital after n terms)
We then find the very important compound interest formula. Percentage and interest 3
n (1.4) Kn = K(1+r)
Kn is the accumulated capital, when the capital has gained interest in n terms with an interest rate r.
The compound interest formula is by no way limited to capital, but may be applied to any quantity which has a constant percentage growth.
If a quantity b has a percentage rate in one period it has grown to bn,, after n-periods where:
n (1.5) bn =b(1+r)
If this is perceived as a function, we may write: f(n) = b(1+r)n.
Traditionally we put: a = 1+r. a may then be called the projection factor. Thus we may write:
f(n) = b(1+r)n f(n) = ban where a = 1+r r = a - 1
The interface between the compound interest formula and the exponential functions f(x) = b ax should then be obvious, since we only have to replace the integer variable n with the real variable x.
2.1 Eksempel. 1. A deposit in a bank having an interest rate of 4.5% p.a is 2.000. The term length is 6 months. Find the deposit after 7 years. The number of terms are then n =14. The interest rate is r = 0.0225 and K = 2000. When we insert in (2.1), we find:
14 K14 = 2000 (1,0225) = 2.730.97
2. At a purchase on installment, the buyer is offered a loan having monthly terms at an interest rate of 1.5%. You may think that this should amount to: 12∙ 1.5% = 18% per annum? Calculate the true effective annual interest. We apply (2.1) with k = 1, to se how much 1 will grow to in 12 terms with an interest rate of 1.5% =0.015.
12 k12 = (1.015) = 1.1956
Which correspond to a yearly interest rate of 19.56%.
Because of the interest rate on the interest, the effective interest rate in n terms will be somewhat larger than n times the interest rate in 1 term.
3. The stock of herring in the Baltic see in mill. tons, could since 1987 be described by an exponential function f(x) = 132(0,82)x, where x is the number of years passed since 1987. Determine the yearly rate of growth. a = 0.82, so r = 0.82-1 = -0.18 = - 18%.
The stock of herring in the Baltic see has thus decreased yearly by 18% since 1987.
Percentage and interest 4
3. The average rate of growth The average rate of growth is defined as the constant rate which gives the same growth in the same period.
The average rate of growth is best illustrated by an example.
Example The growth in the production of a company was 2% in the period 1980-82, It was 5.5% in the period 1982-85, 1.5% in the period 1985-86, -4.3% in the period 1986-88 and -1.8% in the period 1988 - 1990. What is the average growth in this 10 years period? First we calculate the actually growth in the 10 years period. According to the formula (1.5), we have:
2 3 2 2 2 3 2 2 (1+r1) (1+r2) (1+r3)(1+r4) (1+r5) = (1.02) (1.055) (1.015)(0.957) (0.98) = 1.0907.
By the average rate of growth, we understand the constant rate of growth that gives the same growth in the same period. Thus we have:
10 2 3 2 2 10 (1+r) = (1+r1) (1+r2) (1+r3)(1+r4) (1+r5) <=> (1+r) = 1.0907
(1 r ) 10 1.0907 1.0087 r 0.0087 0.87 % (The average rate of growth)
The same procedure can be applied in any other example. One might write a general formula, but it is actually easier to use the example above as the framework for another specific application. Savings annuities and debts annuities 5
Chapter 2. Savings annuities and debts annuities
1. Annuities A savings annuity is a series of deposits, which is made with a fixed period. The period between two periods is called a term. The amount deposed is due to an interest rate r.
To insure the correctness of the formulas derived below, the deposits must be done simultaneously with that the interest is ascribed. In the following n is the number of deposits, and An is the value of the total deposits including the ascribed interests. This will correspond to n -1 terms. The constant deposit at each term is y.
To extrapolate a capital one term, we shall as usual multiply by (1+r). Below is written the value of the deposits after 0 terms, 1 term, 2. terms,…, after n-1 terms.
A1 = y A2 = y(1+r) + y 2 A3 = (y(1+r) + y)(1+r) + y = y(1+r) + y(1+r) + y ... n-1 n-2 An = y(1+r) + y(1+r) + ... +y(1+r) + y
If we write the terms in the last expression in the reverse order.
n-2 n-1 An = y + y(1+r) + ... + y(1+r) + y(1+r)
Notice that the nth term in the series (after n deposits) has the exponent n-1 and not n. Further notice that we come from one term to the subsequent term by multiplying with (1+ r).
A series where the next term is obtained by multiplying by the same constant factor is called a geometric series. What we wish is therefore a formula for the sum of a geometric series.
1.1 Formula for the sum of a geometric series. We shall then write a general geometric series, where the first term is a and the quotient is q.
The sum of the first n terms is denoted Sn.
2 3 n-2 n-1 Sn = a + aq + aq + aq + ... + aq + aq
To derive the formula, we shall first multiply by q , and subsequently subtract Sn from qSn.
2 3 n-1 n qSn = aq + aq + aq + ... + aq + aq 2 3 n-1 n 2 3 n-2 n-1 qSn - Sn = aq + aq + aq + ... + aq + aq - ( a + aq + aq + aq + ... + aq + aq )
We can then see that by the subtraction all the terms aq, aq2 aq3.....aqn-1 cancel, and we find:
Savings annuities and debts annuities 6
n n qSn - Sn = aq -a Sn(q-1) = a(q -1)
And the formula for Sn becomes:
qn 1 (1.1) S a n q 1
Notice that n is number of terms.
1.2 Savings annuity We shall then apply this formula on our savings annuity. We shall only let n be the number of deposits, replace a by y (the deposit) and q by 1+r. We then find:
(1 r)n 1 (1.2) A y n r
Notice that n is number of terms, which is the same as the number of deposits.
1.3 Debts annuities A debt annuity is a loan, which is repaid with a constant amount each term, where some of it is repayment and some of it is interest. Most house mortgages are administrated this way.
In the beginning, most of the payment is interest, but as the loan is repaid, most of the payment is repayment.
We wish to derive a formula for the remaining size of the loan after n terms. The initial amount of the loan is called the principal and is denoted G. The remainder of the loan after n terms (the same as after n in payments) is denoted Gn. The interest rate is r, and the periodic payment is y.
We the write the size of the remaining loan after 0, 1, 2.., n terms
G0 = G G1 = G(1+r) - y ; (The principal loan G has grown to G(1+r), when there is repaid y) 2 G2 = G1 (1+r) - y = (G(1+r)-y))(1+r) - y = G(1+r) - y(1+r) - y 3 2 G3 = G2 (1+r) - y = G(1+r) - y(1+r) - y(1+r) - y ... n n-1 n-2 Gn = G(1+r) - y(1+r) - y(1+r) - ... - y(1+r) - y
What we see, is that we have the following equation.
n Gn = G(1+r) – An .
Where An is the deposit annuity, we derived above.
Savings annuities and debts annuities 7
(1 r)n 1 (1.3) G G(1 r)n y n r
Using this formula, we may calculate the remainder of a debts annuity after n terms. Especially we are interested in the case, where Gn = 0, meaning that the dept has been repaid, including interest.
(1 r)n 1 0 G(1 r)n y (1.4) r 1 (1 r)n G y r To obtain the last expression we have divided by (1+r)n and moved G to the other side.
The last formula is the relation between the principal loan G, the payment y at each term, the interest rate r and the maturity n of the loan. (n = number of terms = number of payments y)
1.5 Example. Repayment of a debt annuity. Let us assume that buyer of a house agrees a 20 year annuity loan of 1,000,000, with monthly payments. The interest rate is 2.4% p.a. 0.2% per month. Calculate the monthly payment y.
Solution: We use the formula (1.4) where we isolate y.
rG 0.002 1,000,000 y 5,240.00 1 (1 r)n 11.002240
Exponential functions 8
Chapter 3. Exponential functions
1. The concept of a power of a number n The symbol a , where a R and n Z is defined by:
a n a a a a (n-factors) an is called a power of a. The number a is called the radix or the root, and n is called the exponent.
1.1 Calculation with powers For powers we have the following important rules:
If a,b R and n,m Z then we have:
1. You may multiply two powers having the same radix, by adding the exponents.
a n a m a nm
2. You may divide two powers with the same radix by subtracting the exponent of the denominator from the exponent of the nominator.
an If n > m: anm a m
3. You may raise a power to a power, by multiplying the exponents.
(a m )n a mn
4. You may raise a product to a power, by raising each factor to the same power.
(a b)n a n bn
5. You may raise a fraction to a power, by raising each of the nominator and the denominator to the same power
n a a n b bn Proof of 1.
n m nm a a aa a a aa a a aa a a a n factors m factors nm factors
Exponential functions 9
Proof of 2.
n factors n a a a a a nm m a a aa a a m factors Proof of 3.
m n m m m m mn (a ) a a aa aa a a a n factors nm factors
Proof of 4.
n n n (a b) ab ab ab ab aa a a bb bb a b n factors n factors n factors
We omit the proof for 5, since it is quite similar.
2. Extension of powers to negative exponents If we wish to extend powers to have zero and negative exponents, then we must obviously claim that the rules 1-5 are still operational without restrictions. First we shall do an analysis, applying the rules 1-5 to determine the significance of e.g. 20 and 3-5.
First we shall look at a0, where a is real number different from zero, applying the rule (1.2):
a 1 a11 a0 . a
We can from this conclude that if (1.2) should still be valid, then we must set:
a0=1 (For all non zero a)
Next we look at: 0 1 a 0n n a a . an an
We then realize that if the second power rule should be valid, then we must set:
n 1 a an
Strictly speaking we need to establish that all the five power rules are still valid with the extension of the exponents to zero and negative integral numbers. If that was not the case the definitions would be meaningless. However, we just settle for a couple of examples, where m and n are positive integers. Exponential functions 10
a n a m a nm (According to the original power rules)
a n a n a m a nm (According to the extension to negative exponents) a m
(a m )n (a m )n a mn (According to the original power rules)
m n 1 1 mn (a ) n mn a (According to the original power rules) a m a
3. Extension of powers to rational number exponents n We remind you of the significance of the symbol: a , where a R \ R - and n Z .
(1.8) n a is the non-negative number when being raised to the power n gives a.
n a b b 0 bn a
Examples. 3 8 2 because 2 0 and 23 8
4 81 3 because 3 0 and 34 81
Notice that in the last example b0 is necessary since both 34 = 81 and (-3)4 = 81.
p q We shall then try to establish a unique definition of the symbol: a , where p,q Z .
The conditions for extending the powers to rational number exponents are as previously that the calculation rules 1 – 5 should still be valid. In the same manner as we did before we make an analysis, using these rule to reveal the (possible) significance of rational number exponents. 1 First we look at a q , where the exponent is a pure fraction.
1 q 1 q a q a q a1 a (3. rule) q On the other hand we have from the definition of n a : q a a , So we have established that: 1 a q q a And we continue:
Exponential functions 11
p 1 p p a q a q q a
Strictly speaking we need to establish that all the five power rules are still valid with the extension of the exponents to rational numbers. This is in principle straightforward, but also trivial, so we skip it. For negative rational exponents, we have that:
p p p q q 0 q 1 a a a 1 a p a q
Notice a useful rewriting:
p p 1 1 q p q a q a q a a p q a p
Example: 1 3 3 1 3 2 1 3 2 5 3 4 3 4 5 2 ( 2) 2 2 3 4 3 3 4 3 12 12 4 2 2 2 2 2 2 62 4 2 6
To extend powers to real exponents is, however, is not so straightforward since it requires knowledge of how to extend the rational numbers to the real numbers. So we shall bypass it, but notice that any real number can be found as the limiting value of an infinite sequence of rational numbers. Said in another way: for any real number you may find a rational number which lies ”infinitely close” to the real number. So we may safely assume that the calculation rules for powers are still valid also for real numbers. It is, however, not possible to reveal the value of the numbers 3 2 or 2 from the calculation rules of powers.
We are then able to define a power for any positive real radix and any real exponent
x a where a R and x R
How to calculate ax, when x is irrational, we must preliminary postpone until we have introduced the logarithmic function. The value of ax can of course be found at any even small mathematical pocket calculator.
When ax is perceived as a function of x, it is called an exponential function with base a, and is written: x f (x) a where a R and x R Exponential functions 12
The power rules apply for all powers with a real exponents x and y. Since these rules are extremely important, we repeat them below in their general form.
a x a x a y a x y a x y (a x ) y a xy a y
x a a x (a b) x a x b x b b x
The exponential function ax is increasing for a >1 and decreasing for 0 < a < 1.
Proof:
A function f is increasing in an interval I, if we for all x1, x2 I have:
x1 x2 f (x1 ) f (x2 ) First we note that: x 0 a 1 a x 1 and x 0 0 a 1 a x 1
It then follows: that ax is increasing for a > 1, since:
x2 x2 x1 a x1 x2 x1 x2 x2 x1 0 => a 1 1 a a a x1 And ax is decreasing for a < 1, since:
x2 x2 x1 a x1 x2 x1 x2 x2 x1 0 => a 1 1 a a a x1 We state without proof, (since it requires knowledge of logarithms) that:
When a 1: a x for x and a x 0 for x When 0 a 1: a x 0 for x and a x for x x 1 In the figure below is drawn the graphs for f (x) 2x and f (x) . 2 x 1 x You should notice that the graphs are symmetric about the y-axis since: 2 2
All exponential functions have the same characteristic shape.
Exponential functions 13
Logarithmic functions 14
Chapter 4. Logarithmic functions.
1. Logarithmic functions
Since f(x) = ax for a ≠ 1 is monotone (increasing/decreasing), then it has an inverse function. The inverse functions to the exponential functions f(x) = ax are called logarithmic functions with radix a. The inverse function to f(x) = ax is written g(x) f 1(x) . In the case of exponential functions:
1 (2.1) f (x) loga (x)
About a function and the inverse function, we have the relations:
(2.2) y = f(x) x = f -1(y)
Dm(f -1) = Vm(f) and Vm(f -1) = Dm(f)
Applying this to: f(x) = ax, we have:
x y y = a x = loga(y) and y = loga(x) x = a
Dm(loga)= R+ and Vm(loga)= R
Especially you get :
0 1 a =1 loga(1) = 0 and a =a loga(a) = 1
x The graphs for y = a and x = loga(y) are identical, since the two expressions are equivalent, but when we swap x and y it means mirroring in the line x = y, such that the graph for y = loga(x) is the mirror image of y = ax in the line y = x.
It shows up, as demonstrated below with an example, that all logarithmic functions are proportional to each other. We shall only be occupied with two logarithmic functions. Namely the one with radix 10, called the decimal logarithmic function which is written log, and which has been applied for numerical calculation since the 1600 millennium and up to the invention of digital computers. The other is the theoretically founded logarithmic function, which is called the natural logarithm, and is written ln. It has the radix e= 2,718281828….The number e is an irrational (actually transcendent like pi). We postpone the explanation to this to end of the integral calculus.
The exponential function with radix e is written ex. The four functions ln(x), ex, log(x) and 10x are found on pocket computers. As it apply for all functions, and their inverse functions and for any x.
y = ex x = ln(y) and y = 10x x = log(y)
All logarithmic functions are due to the same calculation rules, so we shall settle for proving the calculation rules for the natural logarithm ln, since all logarithmic functions are proportional. Logarithmic functions 15
For all a,b R and x, y R , the important calculation rules are:
ln(ab) ln(a) ln(b) a ln( ) ln(a) ln(b) b ln(a x ) xln(a)
Proving that: ln(ab) ln(a) ln(b) . We have:
x = ln(a) a = ex and y = ln(b) b = ey. From which we get ab e x e y e x y x y ln(ab) ln(ab) ln(a) ln(b)
a Proving that: ln( ) ln(a) ln(b) b a a a ln(a) ln( b) ln( ) ln(b) ln( ) ln(a) ln(b) b b b
Proving that: ln(a x ) xln(a) .
Setting: y = ln(a) a = ey
a x (e y ) x e yx x y ln(a x ) ln(a x ) xln(a)
For the special case where n is a positive integer and a is a positive real number.
1 1 ln(n a) ln(a n ) ln(a) n
Furthermore we have the important identities, which are often used:
ln(e x ) x eln x x log(10x ) x 10log x x
All logarithmic functions are proportional. We settle for showing it with an example, using log(x) and ln(x).
Using the calculation rules for logarithms, we may write:
ln(x) x 10log x ln(x) ln(10log x ) log x ln(10) log x ln(10) ln(10) = 2.302585… Below is shown the graphs for ln(x), log(x), ex og 10x.
Logarithmic functions 16
Example. Below is proven a somewhat surprising theorem, which shows that an infinite sequence of rational numbers as a limit may have a transcendent number.
1 n (1 ) e for n n
1 n First we show that: ln(1 ) 1 for n . Since ln is continuous and injective (one-to-one), it follows that the n number itself will converge to e.
1 ln(1 ) 1 n 1 ln(1 ) n ln(1 ) n n n 1 n 1 If we put h = such that h 0 for n , and use that ln 1 = 0 , n f (1 h) f (1) we find, since: f '(1) for h 0 , we get h
ln(1h)ln1 1 for h 0 h
Since ln is differentiable in 1 with the differential quotient 1, which proves the theorem. Exponential growth and power growth 17
Chapter 5. Eksponential growth and power growth
1. Exponential growth An exponential growth is a function that is proportional to a exponential function.
If a,b R then an exponential growth is defined as:
f(x) = b∙ax
Exponential growth (or exponential decrease) appears in many places in Science, physics, biology, economy and social science. Often one encounters the task to determine the two real numbers a and b, such that the exponential passes through the two points (x1,y1) and (x2,y2). The method is most conveniently illustrated by an example.
Example. We shall determine the exponential function, passing through the points (-2, ½) and (3,4). We have thus the following equations:
f( 3) = 4 ba3 = 4 and f(-2) = ½ ba-2 = ½
By dividing the last equation by the first:
ba3 4 a5 8 a 5 8 2 1 ba 2
This can then be inserted in one of the two equations to determine b.
3 3 4 ba3 4 b(5 8)3 4 b 4 8 5 f (x) 48 5 (5 8) x (5 8)3
For all exponential functions, we have: f(0) = ba0 = b, so that: y = bax intersects the y-axis in b.
1.1 Solution to exponential equations. An equation of the form f(x) = c bax = c, is always solved, by isolating the exponential function and taking the logarithm on both sides. c log( ) c c c ba x c a x loga x log( ) xloga log( ) x b b b b loga
In this case we have used the decimal logarithm, but we might as well have applied the natural logarithm ln.
Example: 7 ln( ) x 7 53 7 x ln3 ln( ) x 5 0.3063 5 ln 3 Exponential growth and power growth 18
2. Doubling constant and half-live constant Exponential functions have the property, that they have a constant relative (percent) growth for a constant increment of the independent variable. n If we compare an exponential function with the formulas for compound growth, Kn =K(1+r) replacing Kn by b, (1+r) by a and n by x, then it becomes obvious that
n x Kn =K(1+r) is the same as f(x) = ba r is the interest rate and a is called the projection factor. But we may also show the assertion about constant percentage growth directly.
If we make an increment h on x, then we shall first calculate the absolute growth on f(x) and subsequently the relative growth.
Absolute growth: f(x+h) – f(x) = bax+h - bax = bax(ah - 1), and then the relative growth
f (x h) f (x) a h 1 f (x)
Where, we see that the relative growth is independent of x.
This is in contrast to the case of linear growth:
f(x) = ax + b, where f(x+h) – f(x) = ah (independent of x, whereas the relative growth is dependent on x)
For an increasing exponential function, we define the doubling constant T2 as the increment we shall give the independent variable x, so that the function is doubled.
Since an exponential function is characterized, by having constant relative growth, we know in advance that the doubling constant cannot depend on x, since a doubling corresponds to a relative increase on 100%.
If we denote the doubling constant by T2, we must have:
f (x h) f (x) ln 2 log2 aT2 11 aT2 2 T f (x) 2 ln a loga
Correspondingly, we define for a decreasing function the half-life constant T1 as the increment you 2 should give the independent variable to halve the function. 1 Taking the half of a function is the same as a relative growth of -50% = 2 . Following the same procedure as we did above, we find. Exponential growth and power growth 19
1 1 T T ln log f (x h) f (x) 1 1 1 1 a 2 1 a 2 T 2 2 1 f (x) 2 2 2 ln a log a
3. Logaritmic scale.
Above is shown a segment [0,1] of a number line. Below the line is the ordinary coordinate, which we in this case mark as x’. Above is marked the number x, of which the logarithm is x’. Specifically: x’ = log(x). E.g. Above x’ = 0.3010 is placed 3, because log(3)= 0.3010. x is called the logarithmic coordinate, so the logarithm to the logarithmic coordinate is the ordinary coordinate. A number line, where only the logarithmic coordinates are marked is called a logarithmic scale.
If we continue the logarithmic scale from 10 until 100, then it will be an ordinary scale in the interval [1,2] and the division of the scale will be the same. Below we use the logarithmic coordinate 30 as an example.
log(30) = log(10∙3) = log(3)+log(10) = log(3) +1.
When the logarithmic coordinates are multiplied by 10, The result is a displacement of +1 in the ordinary coordinates. Correspondingly with 0.3: log(0.3) = log(3/10)= log(3) -1. So dividing the logarithmic coordinates by 10, it corresponds to a displacement of -1 in the ordinary coordinates.
Below is shown some mappings of logarithmic coordinates into ordinary coordinates.
[1/100,1/10] -> [-2,-1]; [1/10,1] -> [-1,0]; [1,10] -> [0,1]; [10,100] -> [1,2];
A section of the logarithmic scale e.g. [1/100,1/10] or [1,10], is called a decade.
Before the era of the electronic pockets computers (about 1975) logarithmic scales were applied to perform multiplications, divisions and root extraction, sine and cosine with decimal numbers. The instrument to do so, if not by hand, was a so called slide ruler.
It consisted of two (or more) logarithmic scales, but where two of the scale could be displaced relatively another. They were called the stick and the tongue. On the slide ruler was also a slider, apt to fix and read the numbers on various scales. See the figure below.
Exponential growth and power growth 20
If you want to multiply two numbers a and b with a slide ruler, then you use the tongue, so that the number 1 on the tongue is right above a on the stick. Then you move the transparent glider so that the thin reading line is at b on the tongue. On the stick on the reading line you may then read a∙b, since the distances (in ordinary coordinates) to a and b are log(a) and log(b) respectively, and on the reading on the stick we have namely the sum: log(a)+log(b) = log(a∙b), which is the point having the logarithmic coordinate a∙b. This is sought illustrated below.
Another advantage of using a coordinate system with one logarithmic axis is that:
An exponential function is mapped in a straight line on such a semi-logarithmic coordinate system.
The fact is that when you draw a smooth curve through some points from empirical observations, then the only curve you may recognize with certainty is a straight line. That an exponential function is mapped into a straight line in a semi- logarithmic coordinate system, is explained below.
y =bax log y = log b + x∙log a y’ = a’∙x’ + b’
The last expression is clearly a straight line in the ordinary coordinates. To determine the exponential function, you may select two points (x1, y1) and (x2, y2) to determine a, and b, as we have demonstrated earlier.
But more frequently you may choose to determine the doubling-constant (or the half-life constant) in the following way: Find two points on the (logarithmic) y-axis, corresponding to a doubling of the function e.g. y1 = 2 and y2 = 4 or y1 = 19 and y2 = 38. Then read the corresponding numbers x1 and x2 on the x-axis. The doubling constant is then:
T2 = x2 - x1
In quite the same manner the half-life constant T1 is determined as the difference x2 - x1 between 2 two x-values, corresponding to halving of the function values, e.g. y1 = 12 and y2 = 6.
log 2 1 log 2 If f (x) ba x then a is determined from the equation: T a 10 T2 2T2 2 log a Exponential growth and power growth 21
x And the exponential function may also be written as: f (x) b2T2 , as can be seen since it passes through the two points: (0, b) and (T2, 2b) log 2 1 log 2 T2 1 T2 In the same manner for a decreasing exponential function: T1 a 10 ( ) 2 log a 2
4. Power functions A power function is given by the expression:
a (5.1) f (x) b x ; x,b R og a R
When a is a positive integer, then the definition set can be extended to R, and when a is a negative integer, then the definition set can be extended to R\{0}. In the general case the power function is defined by the rewriting:
a aln x (5.2) f (x) bx be ; x,b R and a R
We have earlier dealt with some familiar power functions:
1 f (x) 2x2 , f (x) x1, f (x) x4 , x 1 2 4 f (x) 3 x 3x 2 , f (x) 4x 3 . 3 x2
From the expression: f (x) bxa bealn x , we can see that f (x) bxa is growing for a > 0, this because eax and ln(x) are both growing. On the other hand, for a < 0 f (x) bxa bealn x is decreasing, since e-kx , is a decreasing function, when k > 0 .
Notice the difference between the exponential functions f(x) = b∙ax and the power functions f (x) bxa , is that the significance of a and x has been swapped. The similarity between the two kinds of function types is that the calculations for both function types functions are based on the calculation rules for powers.
Example. Determine the power function f (x)bxa , that passes through the points (2,4) and (7,3). We set up the two equations:
f (2) 4 b 2a 4 and f (7) 3 b 7a 3 4 a a ln b2 4 2 4 4 a 3 0.2296 from which we find b 4.691 a 2 a b7 3 7 3 ln 2 7 If we want to investigate whether an empiric material can be described by a power function, we must apply double logarithmic paper, that is, a coordinate system with two logarithmic axes. Exponential growth and power growth 22
It is namely so that the graph of all power functions is a straight line in a double logarithmic coordinate system. If x’ and y’ as previous denote the ordinary coordinates, we have:
y b x a log y logb alog x y' a x'b'
The last expression is a straight line with slope a. The expression for a power function passing through two points is found as in the example above.
If you have the straight line and want to find the slope, you should not use the numbers on the y' coordinate axis, but use a ruler with normal division and measure: a . x' The number b can then be read as the intersection on the y-axis, where x = 1 (x’ = 0)
This follows from: f (x) b xa f (1) b 1a b