# e^(-6k)+4k+1 = 0 , k = ?

How do you get "k" from that equation? Thanks

Update:

Since this thing's capitalizing my "e", the correct form of the equation would be e^(-6k)+4k+1 = 0

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• 8 years ago

There is no solution.

If k = 0 , the equation becomes e^(0) + 4(0) + 1 = 1 + 1 = 2

If k > 0 , all of the terms on the left side are positive, and [ e^(-6k)+4k+1 ] > 0 .

If k < 0 , the power of e, (-6k) , is positive. The exponential term with the positive power grows faster than the 4k term, which is negative when k < 0 . As a result, the entire expression

[ e^(-6k)+4k+1 ] > 0 when k < 0.

Here's a view of part of the graph of f(k) = e^(-6k)+4k+1 :

http://s1164.photobucket.com/albums/q561/iago9/?ac...

And here's a graph showing g(k) = e^(-6k) and h(k) = 4k on the same coordinate axes :

http://s1164.photobucket.com/albums/q561/iago9/?ac...

You can see that when x < 0 , e^(-6k) is larger positive than (4k) is negative , so

[ e^(-6k) + 4k ] > 0 when x < 0 , and therefore [ e^(-6k)+4k+1 ] > 0 when k < 0.

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As to how to algebraically show that the exponential term with the positive power 'grows faster' than the 4k term when k < 0 is a little difficult. It could be done with a limit approach, but I don't 'see' that at this particular moment.