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Published byWilfrid Gilmore Modified over 9 years ago
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Basic Logarithms A way to Undo exponents
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Many things we do in mathematics involve undoing an operation.
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Subtraction is the inverse of addition
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When you were in grade school, you probably learned about subtraction this way. 2 + = 8 7 + = 10
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Then one day your teacher introduced you to a new symbol ─ to undo addition
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3 + = 10 Could be written 10 ─ 3 =
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8 – 2 =
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2 + ? = 8
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8 – 2 = 2 + 6 = 8
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8 – 2 = 6 2 + 6 = 8
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The same could be said about division ÷
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40 ÷ 5 =
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5 x ? = 40
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40 ÷ 5 = 5 x 8 = 40
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40 ÷ 5 = 8 5 x 8 = 40
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Consider √49
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= ?
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? 2 = 49
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= ? 7 2 = 49
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= 7 7 2 = 49
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Exponential Equations: 5 ? = 25
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Exponential Equations: 5 2 = 25
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Logarithmic Form of 5 2 = 25 is log 5 25 = 2
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log 5 25 = ?
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5 ? = 25
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log 5 25 = ? 5 2 = 25
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log 5 25 = 2 5 2 = 25
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Try this one…
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log 7 49 = ?
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7 ? = 49
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log 7 49 = ? 7 2 = 49
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log 7 49 = 2 7 2 = 49
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and this one…
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log 3 27 = ?
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3 ? = 27
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log 3 27 = ? 3 3 = 27
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log 3 27 = 3 3 3 = 27
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Remember your exponent rules? 7 0 = ? 5 0 = ?
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Remember your exponent rules? 7 0 = 1 5 0 = 1
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log 7 1 = ?
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7 ? = 1
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log 7 1 = ? 7 0 = 1
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log 7 1 = 0 7 0 = 1
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Keep going…
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log 3 1 = ?
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3 ? = 1
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log 3 1 = ? 3 0 = 1
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log 3 1 = 0 3 0 = 1
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Remember this? 1/25 = 1/ 5 2 = 5 -2
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log 5 ( )= ?
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5 ? = 1/25 log 5 ( )= ?
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5 -2 = 1/25 log 5 ( )= ?
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5 -2 = 1/25 log 5 ( )= -2
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Try this one…
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log 3 ( )= ?
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3 ? = log 3 ( )= ?
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3 -4 = log 3 ( )= ?
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3 -4 = log 3 ( )= -4
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Let’s learn some new words. When we write log 5 125 5 is called the base 125 is called the argument
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When we write log 2 8 The base is ___ The argument is ___
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When we write log 2 8 The base is 2 The argument is 8
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Back to practice…
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log 10 1000=?
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10 ? =1000
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log 10 1000=? 10 3 =1000
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log 10 1000=3 10 3 =1000
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And another one
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log 10 ( )=?
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10 ? =
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log 10 ( )=? 10 -2 =
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log 10 ( )=-2 10 -2 =
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log 10 is used so much that we leave off the subscript (aka base)
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log 10 100 can be written log 100
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log 10000 =?
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10 ? =10000
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log 10000 =? 10 4 =10000
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log 10000 = 4 10 4 =10000
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And again
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log 10 = ?
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10 ? =10
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log 10 = ? 10 1 =10
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log 10 = 1 10 1 =10
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What about log 33?
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What about log 33? We know 10 1 = 10 and 10 2 = 100 since 10 < 33 < 100 we know log 10 < log 33 < log 100
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Add to log 10 < log 33 < log 100 the fact that log 10 = 1 and log 100 = 2 to get 1 < log 33 < 2
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A calculator can give you an approximation of log 33. Look for the log key to find out… (okay, get it out and try)
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log 33 is approximately 1.51851394
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Guess what log 530 is close to.
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100 < 530 < 1000 so log 100 < log 530 < log 1000 and thus 2 < log 530 < 3
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Your calculator will tell you that log 530 ≈ 2.72427….
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Now for some practice with variables. We’ll be solving for x.
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log 4 16 = x
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log 4 16 = x 4 ? = 16
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log 4 16 = x 4 2 = 16
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log 4 16 = x x=2 4 2 = 16
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Find x in this example.
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log 8 x = 2
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log 8 x = 2 8 2 = ?
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log 8 x = 2 8 2 = 64
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log 8 x = 2 x=64 8 2 = 64
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Find x in this example.
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log x 36 = 2
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log x 36 = 2 x 2 = ?
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log x 36 = 2 x 2 = 36
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log x 36 = 2 x= 6 x 2 = 36
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We need some rules since we want to stay in real number world. Consider log base (argument) = number The base must be > 0 The base cannot be 1 The argument must be > 0
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Why can’t the base be 1? 1 4 =1 1 10 =2 That would mean log 1 1=4 Log 1 1=10 That would be ambiguous, so we just don’t let it happen.
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Why must the argument be > 0? 5 2 =25 and 25 is positive 5 0 =1 and 1 is positive 5 -2 = 1/25 and that’s positive too Since 5 to any power gives us a positive result, the argument has to be a positive number.
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