Logarithms
??
?? LOGARITHM ??
??
Inverse Functions
• Original function: y = kx => f (x) = kx
• Inverse function:

x
x
-1
x = ky => y = => f (x) =
k
k
Example
f (x) = 2x;f -1(x) =

x
2
Inverses Invert

x
g(x) =
2

f (x) = 2x

(

æ xö
f  g (x) = 2 ç ÷ = x
è 2ø

(

2x
g  f (x) =
=x
x

)
)
y =k
Invert

y =k

x

x =k

y

Now solve for y.

x
???????????

?????????????
Enter the Logarithm

()

loga b = c

means

a =b
c

“Jumping Horse Outline”by warszawianka. Found at
http://openclipart.or...
x =k
Solution:

y

y = logk x
()

loga b = c



a =b
c

Implications? Graphs of inverse functions reflect
over line y = x.
Implications
•
•
•
•

Increasing
Asymptote: y-axis => x>0.
loga1=0.
logaa=1.
Summary
Basic relationship:

y = a <=> x = loga y
x
Next
Yes but so what?
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Introduction to Logarithms

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Introduction to Logarithms

  1. 1. Logarithms
  2. 2. ?? ?? LOGARITHM ?? ??
  3. 3. Inverse Functions • Original function: y = kx => f (x) = kx • Inverse function: x x -1 x = ky => y = => f (x) = k k
  4. 4. Example f (x) = 2x;f -1(x) = x 2
  5. 5. Inverses Invert x g(x) = 2 f (x) = 2x ( æ xö f  g (x) = 2 ç ÷ = x è 2ø ( 2x g  f (x) = =x x ) )
  6. 6. y =k Invert y =k x x =k y Now solve for y. x
  7. 7. ??????????? ?????????????
  8. 8. Enter the Logarithm () loga b = c means a =b c “Jumping Horse Outline”by warszawianka. Found at http://openclipart.org/detail/10712/jumping-horse-outline-bywarszawianka-10712. Public domain.
  9. 9. x =k Solution: y y = logk x
  10. 10. () loga b = c  a =b c Implications? Graphs of inverse functions reflect over line y = x.
  11. 11. Implications • • • • Increasing Asymptote: y-axis => x>0. loga1=0. logaa=1.
  12. 12. Summary Basic relationship: y = a <=> x = loga y x
  13. 13. Next Yes but so what?
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