solve for x where x is a real number
4+7lne(^x^2)=2e^4lnx
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
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