Logarithmic and Exponential Functions
5.1 Mathematics of e, ln and log
5.1.6 Important properties of logarithms
As the logarithm is the inverse operation of the power, taking the logarithm of a power can bring the exponent value down onto the equation line and simplify the expression.
We have two very important sets of relationships, which are very useful when using loga- rithms to simplify ‘power’ equations:
ln(ep)= p log(10p)= p [5.8]
and:
ln(pq)= q × ln(p) log(pq)= q × log(p) [5.9] Other useful values and properties of logarithms include the following relationships:
ln(pq)= ln(p) + ln(q) log(pq)= log(p) + log(q) [5.10]
ln p q = ln(p) − ln(q) log p q = log(p) − log(q) [5.11] ln(e)= 1 log(10)= 1 [5.12] ln(1)= 0 log(1)= 0 [5.13] ln(0)= −∞ log(0)= −∞ [5.14] log(2)≈ 0.30 [5.15] ln(x)= 2.30 × log(x) [5.16]
Example 5.3
Check that you understand the following calculations (to 3 sf where appropriate). Using equation [5.8]: ln(ekT)= kT log(1000)= log(103)= 3 log(0.01)= log(10−2)= −2 Using equation [5.9]: ln(83)= 3 × ln(8) = 6.24 log(ekT)= kT × log(e) = 0.434kT Using equation [5.10]: ln(3× 8) = ln(3) + ln(8) = 1.099 + 2.079 = 3.18
log(200)= log(2 × 100) = log(2) + log(100) = 0.301 + 2 = 2.30
Using equation [5.11]:
ln(3/8)= ln(3) − ln(8) = 1.099 − 2.079 = −0.98
log(0.02)= log(2/100) = log(2) − log(100) = 0.301−2 = −1.70
Q5.5
Use equations [5.8] to [5.16] to evaluate the following expressions without using a calculator:
(i) log(10−0.3) Hint: use equation [5.8]
(ii) ln(e0.62) Hint: use equation [5.8] (iii) log(2) Hint: use equation [5.15]
(iv) log(20) Hint: log(20)= log(2 × 10) and use equation [5.10]
(v) log(0.5) Hint: log(0.5)= log(1 ÷ 2) and use equation [5.11]
(vi) ln(1000) Hint: use equation [5.16]
(vii) ln(2) Hint: use equations [5.16] and [5.15] (viii) log(23.1) Hint: use equations [5.9] and [5.15]
Example 5.4
Show that, by taking natural logarithms of both sides, the equation:
Nt = N0ekt can be converted to the form:
ln(Nt)= ln(N0)+ kt
As the base in the equation is e, we will take natural logs of both sides: ln(Nt)= ln(N0× ekt)
then using equation [5.10] for the logarithm of a product: ln(Nt)= ln(N0)+ ln(ekt) and using equation [5.8]:
ln(Nt)= ln(N0)+ kt
5.1.7
Solving ‘power’ equations with logarithms
In solving equations, we need to make the unknown value (e.g.x) the subject of the equation.
When the variablex is in a power term, we need to bring it down onto the equation line by
using [5.7], [5.8] or [5.9].
Examples 5.1 and 5.2 show how we can use equation [5.7] with simple equations involving powers of 10 and e.
With more complex equations we often need to make some initial rearrangements before taking the inverse operation.
Example 5.5
Solve the following equation to findt: Nt = N0ekt
whereN0= 4260, Nt= 520 and k = −0.048 day−1. As the units ofk are ‘per day’ we know that the units of time,t, will be ‘days’.
Swap the equation round to put the t term on the LHS (Rule 2), and substituting
values:
4260× e−0.048×t= 520
Divide both sides by 4260 to leave the ‘e term’ clear on the LHS (Rule 5): e−0.048×t = 520/4260 ⇒ 0.1221(4 sf)
Now we can use the inverse operation for e, Rule 6 (3.4.2), or equation [5.7]: −0.048 × t = ln(0.1221) ⇒ −2.103
and finally divide both sides by −0.048 (Rule 5):
t = −2.103/(−0.048) ⇒ 43.8 days (3 sf)
The calculation shows that the exponential decay will fall from an initial value,
N0= 4260, to a value Nt= 520, after a time t = 43.8 days.
In problems where the base of the power is neither 10 nor e, it is necessary to take logarithms of both sides of the equation and then to use equation [5.9].
Example 5.6
To solve equations of the form 86y = 0.6:
The ‘base’ is neither e nor 10 so we can take either ln (base e) or log (base 10) of both sides.
For example, taking logs (base 10) of both sides gives: log(86y)= log(0.6)
Using equation [5.9] we can write the LHS of the above equation as: log(86y)= 6y × log(8) ⇒ 6y × 0.903
Hence combining the above two equations:
6y× 0.903 = log(0.6) ⇒ −0.2218
giving:
Q5.6
Solve each of the following equations (i.e. find the value ofp):
(i) 2e3p= 22 (iii) 4p= 33
(ii) 1.2× 102p= 18 (iv) 0.3= 0.2p
5.1.8
Using logarithms to linearize ‘power’ equations
It is sometimes possible to use linear regression for problems involving power equations by following these steps:
1. Take logarithms of both sides to bring all the powers onto the equation line. 2. Choose appropriate variables (4.3.3) to plot the equation as a straight line. 3. Perform a linear regression to calculate ‘best-fit’ slope and intercept. 4. Interpret the unknown values from the slope and intercept.
Example 5.7
A student believes that the period of swing, T , of a simple pendulum is related to its
length,L, by a power equation of the form: T = k × Ln
wherek is an unknown constant and n is an unknown constant power.
The student records the following periods,T , for different pendulum lengths, L.
L(m) 0.20 0.40 0.60 0.80 1.00
T (s) 0.89 1.25 1.65 1.71 1.98
The calculation for the best estimates for the values of k and n is given in the following text .
The data forT and L recorded in Example 5.7 does not give a straight line, so the first step
is to linearize the equationT = k × Ln, by taking logarithms of both sides:
log(T )= log(k × Ln)⇒ log(k) + log(Ln)
Using equation [5.9] to bring out the power,n, from log(Ln):
log(T )= log(k) + n × log(L)
If we compare this equation with:
y= c + m × x
we can see that log(T ) is equivalent to y and log(L) is equivalent to x.
Plotting log(T ) on the y-axis against log(L) on the x-axis:
log(L) −0.6990 −0.3979 −0.2219 −0.0970 0.0000
log(T ) −0.0506 0.0969 0.2175 0.2330 0.2967
should give a straight line (4.3.2) with a slopem= n and an intercept c = log(k).
The calculation for Example 5.7 is performed on the Website using Excel, and a linear regression analysis gives values for:
Slope, m= 0.4944
Intercept, c= 0.2987
From the slope we calculaten directly: n(= m) = 0.4944
From the intercept we know that log(k)(= c) = 0.2987
Taking the inverse of the log we calculate that:
k= 100.2987⇒ 1.989
This gives a best-estimate equation:
In fact the true equation has values (g is the acceleration due to gravity): T = 2π L g ⇒ 2.006 × L 0.5
Note that it would be possible to use natural logarithms for the above calculation. In this case the value ofn would still equal the slope, n= m, but the value of k would be calculated using k= ec.
Q5.7
For each of the following equations, identify the functions that would be used for each axis to plot the equation as a straight line. In each case, identify how the unknown power can be determined from the regression line.
(i) V = pAk V is the volume of an animal of surface area A (p and k are unknown constants)
(ii) E= σ (T +273)z E is the intensity of radiation emitted from an object at a
temperatureT◦C (z and σ are unknown constants)
5.1.9
Logarithmic scales
Many systems in science are measured as a ratio of values. For example, in microbiology a ‘1-log’ decrease in bacterial population is used to indicate a drop in population by a factor of 10 to one-tenth of the initial value. Similarly a ‘2-log’ decrease would be a drop in population to one-hundredth of the initial value etc.
Q5.8
A bacterial population has an initial value of 5.0× 107. Calculate the population following:
(i) a ‘1-log’ decrease (ii) a ‘3-log’ decrease
P , of the sound with the power density, P0 = 1.0 × 10−12W m−2, which is the quietest sound that can just be heard in normal hearing:
Loudness, L(dB)= 10 log(P /P0) [5.17]
When comparing two sounds the difference in loudness,L1− L2, will be given by:
L1− L2= 10 log(P1/P2) dB [5.18]
The difference in loudness depends on the ratio of the power densities, P1 and P2. As
P1 and P2 have the same units, the ratio has no units, and 10 log(P1/P2) is simply a number.
Example 5.8
If the power density of sound is doubled,P1/P2 = 2, calculate the increase in loudness. Difference in loudness:L1− L2= 10 log(2) = 10 × 0.30 = 3.0 dB
A doubling of power gives an addition of 3 dB.
Q5.9
A sound has an initial loudness of 70 dB. Calculate the new loudness using [5.18] if the power density of the sound is:
(i) doubled (i.e. by a factor of 2) (iv) halved
(ii) increased by a factor of 8 (v) reduced by a factor of 8
Another important area where students will encounter the use of logarithms is in electro- chemistry.
The acidity of a solution depends on the concentration of hydrogen ions [H+] expressed in units of mol L−1 (or mol dm−3). However, the acidity is normally measured using the logarithmic pH scale (to base 10), by taking the log of the numerical value of [H+]:
pH= −log([H+]) [5.19]
Example 5.9 compares the logarithmic pH scale with the direct concentration scale.
Example 5.9
Values of [H+] and pH for pure water and example values for a strong acid and a strong base:
Strong acid Pure water Strong base
(example) (example) Hydrogen ion concentration: [H+]/mol L−1 0.1 1.0× 10−7 1.0× 10−13 pH value: pH = − log[H+] = − log(0.1) = 1.0 = − log(1.0 × 10 −7) = 7.0 = − log(1.0 × 10 −13) = 13.0
The logarithmic pH scale can be considered to be a linear scale in ‘acidity’; the pH-value increases for decreasing acidity. The numbers 1.0, 7.0, 13.0 are a more convenient linear representation of ‘acidity’ than the absolute scale of hydrogen ion concentrations for the same ‘acidities’: 0.1, 10−7, 10−13.
Q5.10
Calculate the pH-values equivalent to the following values for [H+]: (i) [H+]= 3.4 × 10−9mol L−1 (ii) [H+]= 3.4 × 10−4 mol L−1
In spectrophotometric measurements, the absorbance,A, of a solution is related to its per-
centage transmittance,T %, according to the equation:
A= − log T % 100 [5.20]
oped in 3.2 and 3.4.
Example 5.10
(i) Calculate the hydrogen ion concentration, [H+], for a solution with pH= 8.3. (ii) Calculate the transmittance,T %, for an absorbance, A= 0.36.
Substituting in equations [5.19] and [5.20]:
8.3= −log([H+]) 0.36= − log
T %
100
Swap the equations around to put the unknown on the LHS using Rule 2 (see 3.2):
−log([H+])= 8.3 − log T % 100 = 0.36 Change the signs of every term on both sides of the equation using Rule 3:
log([H+])= −8.3 log T % 100 = −0.36 Take the inverse of ‘log’ using Rule 6 (3.4.2) or equation [5.7]:
[H+]= 10−8.3 T %
100 = 10
−0.36
Using a calculator to evaluate 10−8.3 and 10−0.36 (also using Rule 5 forT %):
[H+]= 5.01 × 10−9 mol L−1 T %
100 = 0.4365
T %= 100 × 0.4365 ⇒ 43.7%
Q5.11
Using [5.19], calculate the hydrogen ion concentrations equivalent to the following values for pH:
Q5.12
Using [5.20], calculate the missing values in the following table of equivalent values (∞ = infinite absorbance):
Percentage transmittance,T % 0 1 50
Absorbance,A ∞ 3 1 0