Logs 8

Academic Skills Advice
Logs
A log has a base and a number: e.g.
π‘™π‘œπ‘”3 8
base number
This would be read as: β€œlog to the base 3 of 8”.
The most common base is base 10. If you see a log without the base number (e.g. on
your calculator) then you should assume it is base 10.
e.g. π‘™π‘œπ‘”45
is usually read as β€œlog 45” and means β€œlog to the base 10 of 45”.
What does a log mean?
Examples:
π’π’π’ˆπŸ“ πŸπŸπŸ“ means: β€œwhat power do I need to raise 5 to, to get 125?”
∴ π‘™π‘œπ‘”5 125
=3
(because 53 = 125)
The answer to the log
π’π’π’ˆπŸ πŸ”πŸ’ means:
∴ π‘™π‘œπ‘”2 64
β€œwhat power do I need to raise 2 to, to get 64?”
=6
π’π’π’ˆπŸπŸŽπŸŽ means:
∴ π‘™π‘œπ‘”100
© H Jackson 2011 / ACADEMIC SKILLS
(because 26 = 64)
β€œwhat power do I need to raise 10 to, to get 100?”
=2
(because 102 = 100)
1
Change of base
Most new calculators will accept a log to any base by using the general log key (
Enter base here
)
Enter number
here
If you have an older calculator (without the general log key) and need to find a log with a
different base then you can convert to base 10 using the formula:
π‘™π‘œπ‘”π‘₯ 𝐴 =
π‘™π‘œπ‘”10 𝐴
π‘™π‘œπ‘”10 π‘₯
Remember to do: log of number
divided by log of base.
Example:
ο‚·
Find π’π’π’ˆπŸ’ πŸ‘πŸ“ using the log to the base 10 key.
π‘™π‘œπ‘”4 35 =
π‘™π‘œπ‘”10 35
π‘™π‘œπ‘”10 4
= 𝟐. πŸ“πŸ”
On your calculator you just need
to type: π’π’π’ˆ(πŸ‘πŸ“) ÷ π’π’π’ˆ(πŸ’)
Try finding the logs from the previous examples using your calculator (and either the
general log key or the change of base formula).
What about ln?
Ln is a special log because it’s base is e (e is a number approximately equal to 2.718)
Example:
Ln 7 means β€œwhat power do I need to raise β€˜e’ to, to get 7?”
Ln 7 = 1.946 (3dp) because e1.946 = 7
Remember:
e.g.
e and ln β€œundo” each other because they are β€œopposites”.
ln(𝑒 3 ) = 3
𝑒 𝑙𝑛5 = 5
(because the 𝑙𝑛 and the 𝑒 undo each other (or cancel each other out))
(because the 𝑙𝑛 and the 𝑒 undo each other (or cancel each other out))
© H Jackson 2011 / ACADEMIC SKILLS
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Laws of Logs
There are a few rules/laws related to logs that it is useful to learn and know how to use.
π’π’π’ˆ(𝑨𝑩) = π’π’π’ˆ(𝑨) + π’π’π’ˆ(𝑩)
𝑨
π’π’π’ˆ ( ) = π₯𝐨𝐠(𝑨) βˆ’ π₯𝐨𝐠(𝑩)
𝑩
π’π’π’ˆ(𝑨𝒏 ) = π’π’π’π’ˆ(𝑨)
π’π’π’ˆ(𝟏) = 𝟎 (using ANY base)
Examples:
ο‚·
Write π₯𝐨𝐠(πŸ’) + π₯𝐨𝐠(πŸ“) as a single log.
Using the 1st rule:
ο‚·
Write π₯𝐨𝐠(𝟏𝟐) βˆ’ π₯𝐨𝐠(πŸ‘) as a single log.
Using the 2nd rule:
ο‚·
12
log(12) βˆ’ log(3) = log ( 3 ) = log(4)
Find π’π’π’ˆπŸπŸ“ 𝟏.
Using the 4th rule:
ο‚·
log(4) + log(5) = log(4 × 5) = log(20)
π‘™π‘œπ‘”15 1 = 0
Write 𝟐 π₯𝐨𝐠 πŸ’ (πŸ“) + πŸ‘π₯𝐨𝐠 πŸ’ (πŸ•) as a single log.
We can’t just multiply the 5 and the 7 because of the numbers in front of the logs. We
need to deal with them 1st.
ο‚·
Using the 3rd rule:
2 log 4 (5) + 3 log 4 (7) = log 4 (52 ) + log 4 (73 )
Using the 1st rule:
log 4 (52 ) + log 4 (73 ) = log 4 (52 × 73 ) = log 4 (8575)
Given that π₯𝐨𝐠(πŸ“) β‰ˆ 𝟎. πŸ”πŸ—πŸ—, find π’π’π’ˆ(πŸπŸπŸ“) without using a calculator.
You could use the 1st OR the 3rd rule for this question.
Using the 1st rule:
OR:
Using the 3rd rule:
© H Jackson 2011 / ACADEMIC SKILLS
log(125) = log(5 × 5 × 5) = log(5) + log(5) + log(5)
= 0.699 + 0.699 + 0.699 = 𝟐. πŸŽπŸ—πŸ•
log(125) = log(53 )
= 3log(5)
= 3 × 0.699 = 𝟐. πŸŽπŸ—πŸ•
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