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zordoloom Group TitleBest ResponseYou've already chosen the best response.0
What's the question?
 2 years ago

aceace Group TitleBest ResponseYou've already chosen the best response.0
\[\log_{4}0.125 \] ... evaluate without a calculator
 2 years ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.0
Oh, that's simple. Do you know how to do base change?
 2 years ago

aceace Group TitleBest ResponseYou've already chosen the best response.0
no base change allowed
 2 years ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.0
You divide (log(0.125)/(log(4))
 2 years ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.0
There is really no other way to enter it in a calculator
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
So the log equals some unknown quantity, let's call it x. \[\large \log_{4} 0.125=x\] Do you know how to rewrite this as an exponential?
 2 years ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.0
Oh, sorry, I didn't notice the no calculator allowed. Okay....
 2 years ago

aceace Group TitleBest ResponseYou've already chosen the best response.0
i got it though.... i'll show you the method and tell me if its right
 2 years ago

aceace Group TitleBest ResponseYou've already chosen the best response.0
\[\log_{4} 8\] then \[(\log_{4}2 + \log_{4}4)\] then (0.5 +1) then 1.5
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Hmm yah I suppose that works! :) That's a little different than how I would have done it. But it get's you familiar with log rules, so that's good :D
 2 years ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
0.125 = 1/8 and let \(x = \log_4\dfrac{1}{8}\) Convert it to exponent form, then find x
 2 years ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Yah that was the other way c: hehe Either way though!
 2 years ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
Yeah, just saying.
 2 years ago
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