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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.

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