Basic Logarithms A way to Undo exponents Many
Basic Logarithms A way to Undo exponents
Many things we do in mathematics involve undoing an operation.
Subtraction is the inverse of addition
When you were in grade school, you probably learned about subtraction this way. 2+ =8 7+ = 10
Then one day your teacher introduced you to a new symbol ─ to undo addition
3 + = 10 Could be written 10 ─ 3 =
8– 2= 2+6=8
8– 2=6 2+6=8
The same could be said about division ÷
40 ÷ 5 = 5 x ? = 40
40 ÷ 5 = 5 x 8 = 40
40 ÷ 5 = 8 5 x 8 = 40
Consider √ 49
=? 2 7 = 49
=7 2 7 = 49
Exponential Equations: ? 5 = 25
Exponential Equations: 2 5 = 25
2 Logarithmic Form of 5 = 25 is log 525 = 2
log 525 = ?
log 525 = ? ? 5 = 25
log 525 = ? 2 5 = 25
log 525 = 2 2 5 = 25
Try this one…
log 749 = ?
log 749 = ? ? 7 = 49
log 749 = ? 2 7 = 49
log 749 = 2 2 7 = 49
and this one…
log 327 = ?
log 327 = ? ? 3 = 27
log 327 = ? 3 3 = 27
log 327 = 3 3 3 = 27
Remember your exponent rules? 0 7 =? 0 5 =?
Remember your exponent rules? 0 7 =1 0 5 =1
log 71 = ?
log 71 = ? ? 7 =1
log 71 = ? 0 7 =1
log 71 = 0 0 7 =1
Keep going…
log 31 = ?
log 31 = ? ? 3 =1
log 31 = ? 0 3 =1
log 31 = 0 0 3 =1
Remember this? 2 1/25 = 1/ 5 = 5 -2
log 5( )= ?
log 5( )= ? ? 5 = 1/25
log 5( )= ? -2 5 = 1/25
log 5( )= -2 -2 5 = 1/25
Try this one…
log 3( )= ?
log 3( )= ? -4 3 =
log 3( )= -4 -4 3 =
Let’s learn some new words. When we write log 5 125 5 is called the base 125 is called the argument
When we write log 2 8 The base is ___ The argument is ___
When we write log 2 8 The base is 2 The argument is 8
Back to practice…
log 101000=?
log 101000=? ? 10 =1000
log 101000=? 3 10 =1000
log 101000=3 3 10 =1000
And another one
log 10( )=?
log 10( )=? ? 10 =
log 10( )=? -2 10 =
log 10( )=-2 -2 10 =
log 10 is used so much that we leave off the subscript (aka base)
log 10 100 can be written log 100
log 10000=?
log 10000=? ? 10 =10000
log 10000=? 4 10 =10000
log 10000= 4 4 10 =10000
And again
log 10 = ?
log 10 = ? ? 10 =10
log 10 = ? 1 10 =10
log 10 = 1 1 10 =10
What about log 33?
What about log 33? 1 We know 10 = 10 2 and 10 = 100 since 10 < 33 < 100 we know log 10 < log 33 < log 100
Add to log 10 < log 33 < log 100 the fact that log 10 = 1 and log 100 = 2 to get 1 < log 33 < 2
A calculator can give you an approximation of log 33. Look for the log key to find out… (okay, get it out and try)
log 33 is approximately 1. 51851394
Guess what log 530 is close to.
100 < 530 < 1000 so log 100 < log 530 < log 1000 and thus 2 < log 530 < 3
Your calculator will tell you that log 530 ≈ 2. 72427….
Now for some practice with variables. We’ll be solving for x.
log 416 = x
log 416 = x ? 4 = 16
log 416 = x 2 4 = 16
log 416 = x x=2 2 4 = 16
Find x in this example.
log 8 x = 2
log 8 x = 2 2 8 =?
log 8 x = 2 2 8 = 64
log 8 x = 2 x=64 2 8 = 64
Find x in this example.
logx 36 = 2
logx 36 = 2 2 x =?
logx 36 = 2 2 x = 36
logx 36 = 2 x= 6 2 x = 36
We need some rules since we want to stay in real number world. Consider logbase(argument) = number n The base must be > 0 n The base cannot be 1 n The argument must be > 0 n
Why can’t the base be 1? n n n 14=1 10 1 =2 That would mean ¨ log 11=4 ¨ Log 11=10 n That would be ambiguous, so we just don’t let it happen.
Why must the argument be > 0? 52=25 and 25 is positive 0 n 5 =1 and 1 is positive -2 n 5 = 1/25 and that’s positive too n Since 5 to any power gives us a positive result, the argument has to be a positive number. n
- Slides: 108
