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MasterMindXD

  • 4 years ago

solve 2log(x+11)=(1/2)^x.

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  1. Ishaan94
    • 4 years ago
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    what's log's base?

  2. MasterMindXD
    • 4 years ago
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    10

  3. Ishaan94
    • 4 years ago
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    Oh okay...

  4. Ishaan94
    • 4 years ago
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    \[2\log (x+ 11) = \left( \frac{1}{2}\right)^x\]

  5. Ishaan94
    • 4 years ago
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    that is the question right?

  6. MasterMindXD
    • 4 years ago
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    Yeah, let me show you what I've got so far.

  7. MasterMindXD
    • 4 years ago
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    |dw:1324553869100:dw|

  8. cristiann
    • 4 years ago
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    First observe that x=-1 is a solution. Then use the monotonicity of the functions log and exponent (with subunitary base) to proove that this solution is unique

  9. MasterMindXD
    • 4 years ago
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    How do I continue from this?

  10. cristiann
    • 4 years ago
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    Consider two functions: f1(x)=2log(x+11) and f2(x)=(1/2)^x. At x=-1 they meet for x>-1, f1 increases (is above) and f2 decreases (is below), so they don't meet anymore... for -11<x<-1, f1 is below and f2 is above, so they again don't meet. So, a unique meeting point ....-1 How can you find the point -1? Just by geuessing ... no procedure for finding it ...

  11. MasterMindXD
    • 4 years ago
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    How did you find out that they meet at -1? What is the method you used to figure that out?

  12. cristiann
    • 4 years ago
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    No method ... just by trial and error ... you should always check for some common values ... just to see how it behaves ... this is a prefabricated exercise .... so it should have some easy values to be guessed ... for a real/life situation ... you have to apply numerical methods (which anyway are not sure...) and combine them with some reasoning ...

  13. MasterMindXD
    • 4 years ago
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    Oh okay, I'll try that. Thank you :D

  14. cristiann
    • 4 years ago
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    You are welcome ... :)

  15. arijit.mech
    • 4 years ago
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  16. cristiann
    • 4 years ago
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    Seems to be another equation? (an extra x on the left-hand side?)

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