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algebra regarding logs

log_3 (log 3x)=1. I have another 72 questions similar to this. can you answer this question?

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2 Answers

Chris, you have to clarify the base of second "inner" logarithm. I might assume that bases are the same, and the problem will be:

log3 (log3 3x) = 1 , and according to properties of logarithm [logb b = 1 and log(ab) = log (a) + log (b)]
we have:

log3 3  +  log3 x  = 3

1 + log3 x  =  3 

log3 x = 2 

32 = x 

x = 9 


If no base is specified then it is understood to be base 10

Yes, it is. But how about human factor?

I just was watching international standards (ISO 31), that defines mathematical signs and symbols for use in physical sciences and technology. Sign for logarithm with base 10 is "lg" , and many countries are using this sign. For me it makes more sense, because it's more likely to forget write the base, than forget to write "o" :)
Anyway, our recipient has 2 versions of the answer ...

Interesting. I never seen that notation in any of my text books from college or high school for that matter but I have seen it on another question around here and I thought that was a typo. I learn something new everyday, thank you for sharing that I'll be sure to remember :). Yes he does, hopefully they are both helpful.

You're very welcome, Xavier!
Me too :) is learning something new here ...

In order to answer this problem you need to be able to convert from log form to exponential form. In general if you have loga(b) = c then the exponential form of that would be ac = b.

So in your problem you have log3(log(3x)) = 1. Here a = 3, b = log(3x) and c = 1 so to convert this to exponential you get 31 = log(3x).

We can repeat this process again to get rid of the other log, this time with a = 10 b = 3x and c = 3. Doing so gives you 3x = 103 and x = 103/3