Question 612384: given log b 2= 0.3562, log b 3=0.5646, log b 5=0.8271 and log b 11=1.2323
evaluate:
log b 21
log b 161
log b 2310
Answer by vleith(2983)   (Show Source):
You can put this solution on YOUR website! You need to understand one of the properties of logs in order to do this
See this URL --> http://www.purplemath.com/modules/logrules.htm
Property 1 says logb(mn) = logb(m) + logb(n)
For your problem 1, you are asked for log b 21
That can be written as log b (2*11)
Which is log b 2 + log b 11
You are given the values for both terms, so just add them up.
Same idea for the other two problem
161 = 2 * 3 * 11 So use add those to get the result
2310 = 5 * 3 * 2 * 11 * 7
Since you are not given the log b 7, you can't solve this one with the info provided
|
|
|
