Here's the question you clicked on:
MasterMindXD
solve 2log(x+11)=(1/2)^x.
\[2\log (x+ 11) = \left( \frac{1}{2}\right)^x\]
that is the question right?
Yeah, let me show you what I've got so far.
|dw:1324553869100:dw|
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
How do I continue from this?
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 ...
How did you find out that they meet at -1? What is the method you used to figure that out?
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 ...
Oh okay, I'll try that. Thank you :D
You are welcome ... :)
Seems to be another equation? (an extra x on the left-hand side?)