This is not a necessary part of the course i'm taking, but an interesting problem.

  • Thread starter M4th
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  • #1
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Homework Statement



Let a; b; c [itex]\in[/itex] (1,∞) and m; n [itex]\in[/itex] (0,∞). Prove that

\log_{b^mc^n} a + \log_{c^ma^n} b +\log_{a^mb^n} c \ge \frac 3 {m + n}

Homework Equations





The Attempt at a Solution


I do not even know where to start. A coherent explanation and possible solutions would greatly farther my knowledge of mathematics. Thanks for any and all help.
 
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Answers and Replies

  • #2
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Do you mean [tex]\log_{b^mc^n} a + \log_{c^ma^n} b +\log_{a^mb^n} c \ge \frac 3 {m + n}[/tex]

Or something else?
 
  • #3
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And what does "a; b; c (1;1) and m; n (0;1)" mean?
 
  • #4
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To your first post yes and thank you. Also, Let a; b; c [itex]\in[/itex] (1,∞) and m; n [itex]\in[/itex] (0,∞).
 
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  • #5
SammyS
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Homework Statement



Let a; b; c [itex]\in[/itex] (1,∞) and m; n [itex]\in[/itex] (0,∞). Prove that

[itex]\log_{b^mc^n} a + \log_{c^ma^n} b +\log_{a^mb^n} c \ge \frac 3 {m + n}[/itex]

Homework Equations



The Attempt at a Solution


I do not even know where to start. A coherent explanation and possible solutions would greatly farther my knowledge of mathematics. Thanks for any and all help.
See how you might use the "Change of base" formula:
[itex]\displaystyle
\log_T(P)=\frac{\log_R(P)}{\log_R(T)}=\frac{\log_{10}(P)}{\log_{10}(T)}=\frac{\ln(P)}{\ln(T)}[/itex]​
 
  • #6
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I think you could use this property [tex]\log_b x = \frac {\log_k x} {\log_k b}[/tex] Using that, for example, [tex] \log_{b^mc^n} a = \frac {\log_a a}{\log_a b^mc^n} = \frac {1}{\log_a b^mc^n}[/tex] You could deal with the other logs similarly, so you would get some rational expression involving sums and products of [itex]m\log_a b[/itex] with permutations of a, b, c, m and n. Another consequence of the formula above is that [tex]\log_b a = \frac {\log_a a} {\log_a b} = \frac 1 {\log_a b}[/tex] so you should be able to express everything in terms of [itex]A = \log_a b[/itex], [itex]B = \log_a c[/itex] and [itex]C = \log_b c[/itex] and m and n. After some algebra (probably, quite some algebra) you should be able to simplify that into something manageable and prove the inequality.
 
  • #7
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Well, that certainly clears a lot up for me, but I would very much appreciate some examples of where to go from there. (As you can tell, I'm clearly not versed well enough in mathematics to attempt such a problem). Thank you for any more input and helping me get a better grasp on mathematics.
 
  • #8
SammyS
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M4th,

Where did you get this problem from? What is the level of the mathematics that might be used to solve this problem?

BTW: Notice what you get if a = b = c .
 
  • #9
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I am in a pre-calculus course but my teacher chose me to be part of a separate "group" of students who solves problems outside of the classroom. I would just like to be able to simply understand what is going on in some of these problems.
 
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  • #10
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Using the properties of the logarithm as we already discussed, you can show that [tex]
\\ \log_{b^mc^n} a + \log_{c^ma^n} b + \log_{a^mb^n} c
= \frac {1} {mA + nB} + \frac {1} {mB/A + n/B} + \frac {1} {m/A + nA/B}
[/tex] where [tex]1 \le A \le B[/tex] Then you will need to prove [tex]\frac {1} {mA + nB} + \frac {1} {mB/A + n/B} + \frac {1} {m/A + nA/B} \ge \frac 3 {m + n}[/tex]
 
  • #11
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This is a great forum, I already have learned a bit more about logarithms just watching how you've broken this problem up into its subsequent pieces. I thank you for such rapid responses and hospitality. Im looking forward to learning more about mathematics from you all and would greatly appreciate some more information on this and many more problems to come in my quest to broaden my grasp on this subject. I would also greatly appreciate a worded summary as to how you get to these results. I understand alot of it, but am still missing some peices and any more details would be way more than I expected to get but would also be extremely helpful.

Thank you again
 
  • #12
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This forum is great because it helps you understand the problem and learn how to solve it. Learn, but not just copy a complete solution. That would be against the rules. The rules require that you make an attempt at solving the problem. Here that would be trying to get the expression to the form given above. You have been given enough information to do so. Now you should try and tell us whether that worked and when not, what exactly went wrong.
 
  • #13
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Yes my point exactly. My goal is to be able to arrive at, and transform equations with ease into more workable ones as you have above and really have a good grasp at how to do so with any other similar problems.

Now one problem in my understanding is I grasp how logba=logaa/logab and thus I see where

\\log_{b^mc^n} a = \frac {\log_a a}{\log_a b^mc^n} = \frac {1}{\log_a b^mc^n}

but how is that transformed into the last mathematical statement you made
 
  • #14
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explain how you can transform logb^mc^na = 1/logabmcn into the statement below

Using the properties of the logarithm as we already discussed, you can show that [tex]
\\ \log_{b^mc^n} a + \log_{c^ma^n} b + \log_{a^mb^n} c
= \frac {1} {mA + nB} + \frac {1} {mB/A + n/B} + \frac {1} {m/A + nA/B}
[/tex] where [tex]1 \le A \le B[/tex] Then you will need to prove [tex]\frac {1} {mA + nB} + \frac {1} {mB/A + n/B} + \frac {1} {m/A + nA/B} \ge \frac 3 {m + n}[/tex]
 
  • #15
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391


explain how you can transform logb^mc^na = 1/logabmcn into the statement below
Basic logarithm properties:
[tex]
\log xy = \log x + \log y
\\ \log x^y = y \log x
[/tex]

Apply them to [itex]\log_a b^mc^n[/itex] and see where that gets you.
 
  • #16
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Of course, now it's all starting to come together. Thanks again for your help.
 

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