- #1
MajikWaffle
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Kinda Tricky
Kinda Tricky
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What Is the Solution to This Logarithmic Problem?
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Homework Help Overview
The discussion revolves around a logarithmic problem involving the relationships between variables a and b, specifically expressed through logarithmic equations with different bases. Participants are attempting to navigate the complexities of these relationships.
Discussion Character
- Exploratory, Assumption checking, Problem interpretation
Approaches and Questions Raised
- Participants are exploring the relationships defined by the logarithmic equations and considering how to express a and b in terms of a common variable. There are hints provided, and some participants are questioning the difficulty of the problem.
Discussion Status
Some participants have offered hints and insights, while others express confusion and seek clarification. There is an ongoing exploration of the problem without a clear consensus on the solution.
Contextual Notes
Participants are reminded of the forum's policy against providing direct answers, emphasizing the need for effort in problem-solving. There is also a hint regarding the relationship between the products of certain numbers, which may be relevant to the problem.
- #2
Tide
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Not too tricky - it's just the Golden Ratio.
- #3
JasonRox
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MajikWaffle said:
A nice little question.
I'll give a hint.
9*16=144
12*12=144
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- #4
MajikWaffle
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Yar, still can't get it. Answer anybody 
- #5
Tide
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The answer is in #2 above.
- #6
Avodyne
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The problem is to find [itex]a/b[/itex] given
[tex]\log_9 a = \log_{12}b = \log_{16}(a+b)[/tex]
Let [tex]x = \log_9 a = \log_{12}b = \log_{16}(a+b)[/tex]. Then
[tex]a=9^x,~~~b=12^x,~~~a+b=16^x[/tex]
Now, compare [itex]a(a+b)[/itex] with [itex]b^2[/itex].
[tex]\log_9 a = \log_{12}b = \log_{16}(a+b)[/tex]
Let [tex]x = \log_9 a = \log_{12}b = \log_{16}(a+b)[/tex]. Then
[tex]a=9^x,~~~b=12^x,~~~a+b=16^x[/tex]
Now, compare [itex]a(a+b)[/itex] with [itex]b^2[/itex].
- #7
berkeman
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MajikWaffle said:Yar, still can't get it. Answer anybody
We do not give out answers here at the PF. Show some effort, or your thread will be deleted.
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