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January 3rd, 2019, 08:02 PM   #1
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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
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January 3rd, 2019, 09:00 PM   #2
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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.
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January 3rd, 2019, 09:00 PM   #3
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Quote:
Originally Posted by psaxno View Post
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.
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Last edited by JeffM1; January 3rd, 2019 at 09:04 PM.
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