solve for x where x is a real number

4+7lne(^x^2)=2e^4lnx(Cool)

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- Aug 12th 2009, 12:00 AMRose Wanjohiexponential functions
solve for x where x is a real number

4+7lne(^x^2)=2e^4lnx(Cool) - Aug 12th 2009, 02:25 AMmr fantastic
The equation is not clear. Please re-post it. Ideally, learn some basic latex before re-posting: http://www.mathhelpforum.com/math-he...-tutorial.html

- Aug 12th 2009, 03:43 AMIsomorphism
Guessing it is $\displaystyle 4 + 7 \ln e^{x^2} = 2e^{4\ln x}$

Use the following identities: $\displaystyle 4\ln x = \ln x^4$, $\displaystyle \ln e^{x^2} = x^2$ and $\displaystyle e^{ln x^4} = x^4$

$\displaystyle 4 + 7 \ln e^{x^2} = 2e^{4\ln x} \implies 4 + 7x^2 = 2x^4$

Now observe that $\displaystyle 4 + 7x^2 = 2x^4 $ is a quadratic in $\displaystyle x^2$. Solve the equation and discard the negative root. And then the solution to the original equation is the positive square root of the root.

Can you work out the details and post your answer?

Good luck (Wink)