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Proofs of Properties of Logs

Proofs of Properties of Logs. Thinking through the math (6.11). SAT Prep . Quick poll! 200 is what percent of 20?. SAT Prep . Quick poll! 66 2 + 2(34)(66) + 34 2 =. SAT Prep . Quick poll! 3. Using the properties– rewrite the following into a single argument. log(4) + 3log(2) – log(6)

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Proofs of Properties of Logs

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  1. Proofs of Properties of Logs Thinking through the math (6.11)

  2. SAT Prep Quick poll! • 200 is what percent of 20?

  3. SAT Prep Quick poll! • 662 + 2(34)(66) + 342 =

  4. SAT Prep Quick poll! 3.

  5. Using the properties– rewrite the following into a single argument • log(4) + 3log(2) – log(6) • log(7)2 – log(x) • log(m) + 2log(n)2 + log(5) • log(t)1/3 + log(s)

  6. Using the properties– rewrite the following into a single argument • log(4) + 3log(2) – log(6) = log(4) + log(2)3 – log(6) = log (32/6) = log (16/3) • log(7)2 – log(x) = log(49/x) • log(m) + 2log(n)2 + log(5) = log(m) + log(n4) + log(5) = g(m·n4·5) = log(5mn4) • log(t)1/3 + log(s) = log(st1/3)

  7. Proving property #1–Start logb(xy) = logb x + logb y We start this proof by setting the following: r = logb x and s = logb y Then we rewrite each one: br = x and bs = y What happens if we multiply x and y?

  8. Property #1 Keep going logb(xy) = logb x + logb y r = logb x and s = logb y br = x and bs = y xy = br·bs = br+s If we rewrite the statement: r+s = logb(xy) Finally, undo the substitution: logb x + logb y = logb(xy)

  9. Property #1 Summary logb(xy) = logb x + logb y r = logb x and s = logb y br = x and bs = y xy = br·bs = br+s r+s = logb(xy) logb x + logb y = logb(xy) With this up, can you start the proof for the second one?

  10. Property #2 Start logb (x/y) = logb x – logby We also start this proof by setting the following: r = logb x and s = logb y Then we rewrite each one: br = x and bs = y What happens if we divide x and y?

  11. Property #2 Keep going logb(x/y) = logb x - logb y r = logb x and s = logb y br = x and bs = y x/y = br /bs = br-s If we rewrite the statement: r-s = logb(x/y) Finally, undo the substitution: logb x - logb y = logb(x/y)

  12. Property #2 Summary logb(xy) = logb x + logb y r = logb x and s = logb y br = x and bs = y x/y = br /bs = br-s r-s = logb(x/y) logb x + logb y = logb(xy)

  13. Change of Base Start logbx = log x/ log b We start this one by adding a variable and rewriting it: logbx = y by = x.

  14. Change of Base Keep going If the sides are equal to each other, then the logs of each side must be equal to each other. So, we can log each side: log(by)= log(x) Then bring down that exponent: log(by)= log(x) y·log(b) = log (x)

  15. Change of Base Summary Finally divide: y·log(b) = log (x) y = log (x)/log (b) Since y = logbx at the start, logbx = log (x)/log (b)

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