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Maths and Chemistry for
Biologists
Maths Examples
This section of the course gives worked
examples of the maths covered in Maths 1 and
Maths 2
Powers and units
What is the value of m in terms of n, p and q in
n
p
a
x
a
am =
aq
Remember the rules of powers that when the
numbers are multiplied the powers are added,
when they are divided the powers are subtracted
So am = an x ap x a-q
and m = n + p -q
What is the volume of a muscle fibre of length 5 mm
and radius 10 m expressed in a) m3, b) m3?
(the volume of a cylinder is r2l where r is the radius
and l is the length.  = 3.14)
a) l = 5 mm or 5 x 10-3 m, r = 10 m or 10 x 10-6 m
Volume = 3.14 x (10 x 10-6)2 x 5 x10-3
= 3.14 x 10-10 x 5 x 10-3
= 15.7 x 10-13 m3 or 1.57 x 10-12 m3
b) 1 m = 106 m so 1 m3 = 1018 m3
Hence volume = 15.7 x 10-13 x 1018 = 15.7 x 105 m3
Logs
Use your calculator to evaluate log 5, ln 5,
antilog 0.25 and antiln 0.25
Your calculator will have keys labelled log and ln but
calculators vary as to whether you enter the
number first or press the function key first – if you
do it the wrong way round you get an error.
antilog 0.25 means the number whose log is 0.25.
You get this by using the 10x function which is the
second function of the log key. Similarly for antiln
The answers are 0.6990, 1.6094, 1.7783, 1.2840
Without using a calculator evaluate log 20, log 0.2
and log 8 given that log 2 = 0.3010
For this you need to remember the rules of logs
log 20 = log (2 x 10) = log 2 + log 10 =0.3010 + 1
= 1.3010
log 0.2 = log 2/10 = log 2 – log 10 = 0.3010 – 1
= -0.6990
log 8 = log 23 = 3 x log 2 = 0.9031
(check with your calculator that you got them right)
From the definition of logs show that
ln x = 2.303 log x (only for the enthusiast!)
Let y = ln x then by definition x = ey
Take logs of both sides
log x = log ey = y.log e
log x
Hence y =
= log x
log e
0.4342
Hence ln x = log x = 2.303 log x
0.4342
Equations
Solve the simultaneous equations
3x + 4y = 12
1)
2x + 3y = 8.5
2)
Multiply 1) by 2 and 2) by 3 to get
6x + 8y = 24
3)
6x + 9y = 25.5
4)
Subtract 3) from 4) y = 1.5
Substitute this in 1) 3x + 6 = 12 so x = 2
Solve the equation x2 -3.5x + 1.5 = 0
This is of the from ax2 + bx +c = 0
with a = 1, b = -3.5 and c = 1.5
So
and
3.5  (-3.5)2 - 4 x1x1.5
x
2 x1
3.5  6.25
x
2
Hence x = 3 or x = 0.5
Consider a reaction in which a substance A breaks
down into two products B and C. If the value of the
equilibrium constant is 4 M calculate the
concentrations of A, B ad C at equilibrium with a
starting concentration of A = 1 M
At equilibrium the following condition applies where
[ ] indicate concentrations
[B][C]
K
[A]
If we let the concentrations of B and C at equilibrium
be x M then that of A will be (1 - x) M and so
x.x
4
0.1 - x
contd
This rearranges to 0.4 - 4x = x2 or x2 + 4x - 0.4 = 0
This is of the form ax2 + bx + c = 0 with a = 1, b = 4
and c = -0.4. Applying the usual equation:
- 4  4 - 4 x1x(-0.4)  4  17.6
x

2
2
2
so
- 4  4.1952
- 4 - 4.1952
x
 0.0976 M or x 
 - 4.0976 M
2
2
The concentration cannot be negative so the correct
answer must be that [B] and [C] = 0.0976 M
and [A] = 0.1 - 0.0976 = 0.0024 M
Two parameters x and y are related by
y = aebx
Given the values of x and y below obtain values of a
and b from a plot of ln y against x
x
y
-2
0.74
-1
1.21
0
2.00
1
3.30
2
5.44
contd
The plot shown below was drawn in EXCELTaking
natural logs of the original relationship we get
ln y = ln a + bx
From the trend line equation b = 0.5 and ln a = 0.6931
Hence a = antiln 0.6931 = 2.0
2
1.5
ln y
1
0.5
y = 0.5x + 0.6931
0
-4
-2
-0.5
0
x
2
4