Finding roots of an exponential equation

Thread starter Akash47
Start date Nov 1, 2019
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Exponential Roots

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The discussion focuses on solving the equation involving exponential functions, specifically 3^x and e^x, which cannot equal zero for real x. The only viable equation presented is x^2 - 4 = 0, leading to the solutions x = 2 and x = -2. Participants confirm that this approach is correct without needing additional considerations. The consensus validates the method used to find the roots. The conversation emphasizes the importance of recognizing the properties of exponential functions in the context of solving equations.

Nov 1, 2019

Akash47

Homework Statement: Given that f(x)=(3^x).(e^x).(x^2-4) for real x.Find the root of the function.(setting f(x)=0).

Relevant Equations: No equations required.

I know that both 3^x and e^x can't be 0 for real x. Then x^2-4=0 is the only choice and we get x=2,-2. Am I right? Or should I add something?

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Nov 1, 2019

Delta2

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yes you are right

Nov 1, 2019

Akash47

Thanks

Thread 'Find the polar form of a complex number'

The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.

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