Transcript Slide 1

Dr J Frost
C2 – Logarithms
Starter
Sketch a graph of y = 3x
(Note: this was once used as an exam question)
1 mark: Two of the three criteria.
2 marks: All of the three criteria.
?
1
(a) Correct shape in left quadrant.
(b) Correct shape in right quadrant
(c) y-intercept of 1.
Functions and their inverses
Some operators exist to provide the opposite of others.
Function
4
4
x×3
x+3
4
x2
4
x5
Inverse
12
7
16
1024
x ? 3
4
x ?- 3
4
?
√x
4
5√x
?
4
Functions and their inverses
Some operators exist to provide the opposite of others.
Function
4
3x
Inverse
81
log?3 x
4
How therefore would describe the effect of log3 x in words?
log3 x
We describe this as the “logarithm
of x base 3” or “log of x base 3” or
“taking the log of x base 3”.
It finds the power that, when 3 is raised to it,
gives you x.
i.e. If y = log3 x, then 3y = x
It is the opposite/inverse of exponentiation.
Computing logs
Remember that logarithms find the missing power.
log2 8 = 3?
Bro Tip #1: Imagine what power would slot in
the middle of the two.
log2 8 = 3
3=
log3 9 = 2?
log10 100 = 2?
log4 1 = 0?
log3 3 = 1?
Bro Tip #2: loga 1 = 0 (a > 0)
Bro Tip #3: loga a = 1 (a>0)
Click to start
bromanimation
Computing logs
Remember that logarithms find the missing power.
1
log2( 2 )
1
log2( 8 )
= -1?
= -3?
1
log3( 81)
log4 (-1)
Bro Tip #4: When we take the log of any
value between 0 and 1 (exclusive), we end
up with a negative number.
= -4?
__πi__
= log? 4
e
Bro Tip #5: If you want a real result, you
can only take logs of positive values.
x
0.25
0.58 1
2
y
-2?
-1? 0
? ?
2? 3
2
4
4
1
y = log2 x
8
?
6
4
2
-4
-2
6
8
10
12
14
-2
-4
-6
Click to
brosketch
Using logs
Logs help us solve equations when the power is unknown.
Find the x for which 10x = 500
We can write this as x = log10 500
?
Broculator Tip:
The log button on your calculator is implicitly
base 10. So “[log] [500]” will give you log10 500
More on rewriting Powers as Logarithms
Bro Tip: In both cases, the 2 is the ‘base’.
3
2
=8
log2 8 = 3
= 81
log9 81? = 2
34 = 81
log3 81? = 4
= 55
log3 55? = x
2
9
x
3
Exercises
C2 Chapter 3 Pg 42
Exercise 3B (All questions)
Exercise 3C (Q1, 3, 5, 7)
Laws of Logs
These are 3 laws of logs that you need to remember.
loga xy = loga x + loga y
x
loga y
= loga x - loga y
loga (xk ) = k loga x
Proving these involves rewriting the logs as exponential
expressions, then using laws of indices.
Laws of Logs
Write the following as a single logarithm.
log230 – log26
log2 5?
? 2
2log3 a + log3 b
log3 (a b)
3log4 (a) – 4log4 b
a3
log4 (? b4 )
Put in the form k + loga (..)
3loga (a√b)
?
Laws of Logs
Now the other way round! Write in the form loga x, loga y and loga z.
loga(b2c3)
2loga b +? 3logac
4loga(√b)
2loga b ?
loga(a√b)
1
loga( x
)
1
1
+ 2?loga
-loga x?
b
Using logs in science
The well-known “Moore’s Law” states that the processing power of
computers doubles every two years. It’s been remarkably accurate so far!
If we were to plot the number of transistors against
the year, the type of graph would be exponential.
?
The graph would look rubbish if we chose a range of
values on the y-axis to accommodate all the values,
because except for the last few years, most of the
points would look close to 0.
Using logs in science
The well-known “Moore’s Law” states that the processing power of
computers doubles every two years. It’s been remarkably accurate so far!
This graph gets around the problem by
letting the y-values increase by a factor
of 10 for each unit, rather than
increasing by a constant amount each
time. Technically this is not allowed!
Using logs in science
The well-known “Moore’s Law” states that the processing power of
computers doubles every two years. It’s been remarkably accurate so far!
10
log10(Number of transistors)
9
We could instead use a logarithmic
scale. We can take the log base 10 of
these values. Then we’ll get 3, 4, 5, 6, ...,
which is now allowed!*
8
7
6
5
4
Logarithmic scales turn exponential
graphs into linear ones (i.e. a straight
line), thus making it much easier to plot
all the points together.
3
* Although realistically, a scale of 10, 100, 1000, etc. is permissible as long as we’re mindful that it’s a logarithmic scale.
Using logs in science
Logarithmic scales are used for earthquakes and noise levels.
loga ax = 1 + loga x
From our laws of logs, in base a...
when a quantity gets a times bigger,
the overall result only increases by 1.
Thus using logarithms turns a factor difference into a constant difference.
The Richter Scale is used to measure the magnitude of earthquakes. The scale is
logarithmic (base 10): it means if amplitude of the earthquake’s waves gets 10 times
bigger, the value on the Richter Scale only increases by 1.
Earthquakes of magnitude 6 vs 7 doesn’t look like a substantial difference, but just the
one point difference means it’s ten times worse!
Exercises
C2 Chapter 3 Pg 42
Exercise 3D (All questions)
Solving ax = b
We saw how we can solve equations like 10x = 125.
But what about when the base is different, e.g. 3x = 20?
OPTION 1: The “Look at me, I have a fancy calculator” method
?
x = log3? 20
OPTION 2: The “change of base” method
3x
= 20
log10 3x =? log10 20
x log10 3 =? log10 20
log 20
?
x = log 3
10
10
Super Bro Tip:
Whenever you’re trying to solve
an equation where the variable
appears in the power, your first
instinct should always be TAKE
LOGS DAMMIT!
Changing the Base
We saw a second ago that we could “change the base” to
find log3 in terms of log10.
3x = 20
METHOD 2
METHOD 1
x = log3 20
x=
log10 20
log10 3
More generally, to change the base from a to b:
logb x
logaa x =
logb a
Click to start
bromanimation
All your base belong to us
Express these logarithms in the specified new base.
log2 5 in base 10
log7 10 in base 12
log10 5 in base 9
log5 10 in base 10
log10 5
?
log10 2
log12 10
?
log12 7
Broculator Tip:
This is how you
could find log25
on a calculator if
you didn’t have
the fancy extra
log button. i.e.
Change to base
10!
log9 5
?
log9 10
log10 10
___1___
?
=
log10 5
log10 5
All your base belong to us
Bro Tip: When you switch the argument and base, you take the reciprocal.
___1___
loga b = log a
b
Solving Equations involving Variables in Powers
7x+1 = 3x+2
Variables appear in
powers, so apply Bro Tip.
log 7x+1 = log 3x+2
(x+1)log 7 = (x+2)log 3
(The base of the log doesn’t
matter)
xlog 7 + log 7 = xlog 3 + 2log 3
xlog 7 - xlog 3 = 2log 3 – log 7
x(log 7 - log 3) = 2log 3 – log 7
x=
2log 3 – log 7
log 7 – log 3
Solving Equations involving Variables in Powers
52x + 7(5x) – 30 = 0
(5x)2 + 7(5x) – 30 = 0
Let y = 5x
y2 + 7y – 30 = 0
(y+10)(y-3) = 0
y = -10 or y = 3
5x = -10 or 5x = 3
x = log5(-10) or x = log53
Bro Tip:
By recognising that 52x = (5x)2,
we’ve turned the equation
into a quadratic!
22x + 3(2x) – 4 = 0
x = 0?
3x-1 = 8x+1
?
x = -3.24
Exam Technique
When solving, you can often either:
a) Get in the form logab = c. Then rearrange as ac = b
b) Get in the form logab = logac. Then b = c.
EdExcel exam questions:
Solve log2 (2x + 1) – log2 x = 2
Solve log 2 (x + 1) – log 2 x = log 2 7
Answer: x = 1/2
?
Answer: x = 1/6
?
Exam Technique
When solving, you can often either:
a) Get in the form logab = c. Then rearrange as ac = b
b) Get in the form logab = logac. Then b = c.
Edexcel exam questions:
Solve
log2 32  log2 16
 log2 x
log2 x
Answer: x = 8 or?1/8
log2 (11 – 6x) = 2 log2 (x – 1) + 3
? 3/2
Answer: x = -1/4 or