6.2 Properties of logarithms

${\text{log}}_2\left(2\right)=$

${\text{log}}_2(2*2)=$

${\text{log}}_2(2*2*2)=$

${\text{log}}_2(2*2*2*2)=$

${\text{log}}_2(2*2*2*2*2)=$

Based on exercises #1-5, finish this sentence in words: when you take ${\text{log}}_2$ of a number, you find:

${\text{log}}_2\left(8\right)=$

${\text{log}}_2\left(16\right)=$

${\text{log}}_2(8*16)=$

${\text{log}}_3\left(9\right)=$

${\text{log}}_3\left(27\right)=$

${\text{log}}_3(9*27)=$

Based on exercises #7-12, write an algebraic generalization about logs.

Now, let’s dig more deeply into that one. Rewrite exercises #7-9 so they look like exercises #1-5: that is, so the thing you are taking the log of is written as a power of 2.

• A

  #7:

• B

  #8:

• C

  #9:

• D

  Based on this rewriting, can you explain why your generalization from #13 works?

${\text{log}}_5\left(25\right)=$

${\text{log}}_5\left(\frac{1}{25}\right)=$

${\text{log}}_2\left(32\right)=$

${\text{log}}_2\left(\frac{1}{32}\right)=$

Based on exercises #15-18, write an algebraic generalization about logs.

${\text{log}}_3\left(81\right)=$

${\text{log}}_3(81*81)=$

$\log_{3}81^2$

${\text{log}}_3(81*81*81)=$

$\log_{3}81^3$

${\text{log}}_3(81*81*81*81)=$

$\log_{3}81^4$

Based on exercises #20-26, write an algebraic generalization about logs.