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 January 3rd, 2019, 07:02 PM #1 Newbie   Joined: Jan 2019 From: Europe Posts: 3 Thanks: 0 Solve an equation with two variables Hello, I am trying to solve this equation, but I can't find any way to do that. The solution has to be a=b=0, but how can I prove that? Thank you in advance. 2e^(a^2)+ln(b^2+1)=2 January 3rd, 2019, 08:00 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 If a and b are real, 2e^(a^2) is at least 2 and ln(b^2 + 1) is at least 0. Hence 2e^(a^2) + ln(b^2+1) = 2 implies 2e^(a^2) = 2 and ln(b^2+1) = 0. Solving those equations gives a = 0 and b = 0. If a and b need not be real, there are other solutions. Thanks from topsquark, JeffM1 and psaxno January 3rd, 2019, 08:00 PM   #3
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Quote:
 Originally Posted by psaxno Hello, I am trying to solve this equation, but I can't find any way to do that. The solution has to be a=b=0, but how can I prove that? Thank you in advance. 2e^(a^2)+ln(b^2+1)=2
You can certainly prove that a = 0 = b is A solution through substitution.

$a = 0 = b \implies \{2e \text {^}(a^2)\} + ln(b^2 + 1) = \{2e \text {^} 0^2\} + ln(0 + 1) =$

$(2 * e^0)+ ln(1) = (2 * 1) + 0 = 2.$

That is a proof.

EDIT: skipjack has provided a proof that is the only real solution.

Last edited by JeffM1; January 3rd, 2019 at 08:04 PM. Tags equation, solve, variables Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Indigo28 Geometry 1 September 23rd, 2017 09:32 AM MathAboveMeth Differential Equations 10 December 22nd, 2016 03:09 AM PhizKid Algebra 3 August 2nd, 2012 08:50 AM demipaul Algebra 1 October 3rd, 2009 10:00 PM illuminatisucks Algebra 5 September 25th, 2009 08:11 AM

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