Voicethread: Solving Log Equations I

The first voicethread of this week was a helpful review for me on solving log. equations. The first thing reviewed was solving simple logarithms:

log22 = x

2^x = 2

x=1

 We also reviewed the "smoosh" method:

log(x+1)+log2=3

log((x+1)2)=3

log(2x+2)=3

103 =2x+2 

1000=2x+2

998=2x 

499=x

Finally, we learned something we probably already knew:

log(x+3)=4

x+3=4