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how to find log of log50/15?

Mathematics
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is it (log 50)/15 or log(50/15) ?
\[\log_{50/15} \]

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Other answers:

that has no meaning the base is 50/15 ?
convert 50=5*10 and 15=5*3 so now =log10/3 =log10-log3 =1-log3
now it easily solving i think
lol, no...and she also needs to find log of that.....
waht shold be the value of = -2.303*2*298*log50/15 ? now please elaboarate.
i think -717.69
can't u use calculator ? u can find log (50/15) = log(10/3) using calcy
@ruchi k now can i ask 1 que.?
* log(10/3) using calcy ONLY, not manually
yah @vipul92 hey @hartnn btw the value is -1436 calories.
@hartnn it's ok @vipul92 is just posting answers to people's posts and usually posting wrong answers
-1436 is double the value of expression that u wrote
hey @gaara you don't know how to multiply and subtract two no. and said that i m wrong
excuse me.! @vipul92 this place is for posting questings and answering them, so please don't do useless talks here. that would go against the code of conduct! thanks!
k as u wish
using calculator -2.303*2*298*log(50/15) = -717.697 -1436 is double of that
sorry to everone the question is something like this : -2.303*2*2*298*log(50/15) =
-2.303*2*2*298*log(50/15)=-1436 using calculator
hey i'm asking that how u hav find log in this question
doesn't your calculator have the option to calculate log? if not then u can simplify log(50/15) = log (10/3) = log 10 -log 3 =1-log 3 and u need to know log 3 = 0.477
in xamination calculator r nt preferred.
then i think, u need to remember atleast these 2 : log 2 = 0.301, log 3 = 0.477
hey hw to find log of 20/5?
I'm still trying to figure out the notation. What country are you from?
You can interpret that in different ways: \[\log_{20}{(5)}\] \[\log_{10}(20/5)\] \[\log_{5}(20)\]
If it's \[\log_{10}(20/5) = \log_{10}(4)\] , then it is relatively easy.
But if it's something other than that, then idk
If you want help, you will need to clarify what the notation is.
Okay, so if it is what I think it is, then log(20/5) = 2 log(2) = .602
hey hw u fing it?
\(\log (20/5)=log 4 = log (2^2)=2log2 \\ \text{from the property of log that :} \\ \huge logx^n=nlogx\) and as i mentioned, u should remember log 2 =0.301

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