Question

how would you solve this? (6e^x^2)-e^(7-x^2)=0

Answer 1

move 2nd term to the right \[6 e^{x^2}= e^{7-x^2} \]
then take natural log on both sides:
\[\ln(6)+x^2=7-x^2\]
so \[x^2=\frac{7-\ln(6)}{2}\]
and take sqrt root.... done!

Answer 2

OMG thanks. I didn't think to take the natural log of 6. Thanks =)