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anonymous
 4 years ago
solve 2log(x+11)=(1/2)^x.
anonymous
 4 years ago
solve 2log(x+11)=(1/2)^x.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[2\log (x+ 11) = \left( \frac{1}{2}\right)^x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that is the question right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, let me show you what I've got so far.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1324553869100:dw

cristiann
 4 years ago
Best ResponseYou've already chosen the best response.2First 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How do I continue from this?

cristiann
 4 years ago
Best ResponseYou've already chosen the best response.2Consider 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 ...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How did you find out that they meet at 1? What is the method you used to figure that out?

cristiann
 4 years ago
Best ResponseYou've already chosen the best response.2No 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 ...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh okay, I'll try that. Thank you :D

cristiann
 4 years ago
Best ResponseYou've already chosen the best response.2You are welcome ... :)

cristiann
 4 years ago
Best ResponseYou've already chosen the best response.2Seems to be another equation? (an extra x on the lefthand side?)
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