My Math Forum e^-3.5555x = 0

 Pre-Calculus Pre-Calculus Math Forum

 January 28th, 2019, 02:55 PM #1 Newbie   Joined: Jan 2019 From: Canada Posts: 1 Thanks: 0 e^-3.5555x = 0 It says solve for x. I have no idea how to do this. It is a question on my exponents assignment. Thanks. Maybe I'm completely missing something simple. If you think it's unsolvable at this level, please say so too! Last edited by skipjack; January 28th, 2019 at 09:21 PM.
 January 28th, 2019, 03:07 PM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 Please give the EXACT and complete wording of the exercise. There is no solution to this problem in the real numbers. Thanks from topsquark and leahdm02
January 28th, 2019, 03:24 PM   #3
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Quote:
 Originally Posted by leahdm02 It says solve for x. I have no idea how to do this. It is a question on my exponents assignment. Thanks. Maybe I'm completely missing something simple. If you think it's unsolvable at this level, please say so too!
It's unsolvable at any level since the exponential function is never zero for real or complex exponents.

Last edited by skipjack; January 28th, 2019 at 09:25 PM.

 January 28th, 2019, 03:36 PM #4 Senior Member   Joined: Dec 2015 From: somewhere Posts: 534 Thanks: 81 Do you mean $\displaystyle e^x =-3.5$? $\displaystyle x=\ln(-3.5)\;$ is a complex solution. Thanks from Maschke and topsquark Last edited by skipjack; January 28th, 2019 at 09:17 PM.
January 28th, 2019, 09:15 PM   #5
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 Originally Posted by idontknow Do you mean $\displaystyle e^x =-3.5$? $\displaystyle x=\ln(-3.5)\;$ is a complex solution.
Oh, that's so obvious now that you pointed it out. I'm terrible at figuring out what questions actually mean. Good catch!

Last edited by skipjack; January 28th, 2019 at 09:18 PM.

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