The equation $ e^{2 x}-11 e^{x}+30=0 $ has two solutions. The smaller one is: and the larger one is:

Question

asked Dec 25, 2024 112k views

1 Answer

JensB · Answer 1 · 2024-12-30T15:05:34+0000

Answer:

ln(5) is smaller

ln(6) is larger

Explanation:

$e^(2x)-11e^x+30=0\\u^2-11u+30=0\\(u-5)(u-6)=0\\\\u=5,\,u=6\\e^x=5,\,e^x=6\\x=\ln(5),\ln(6)$

We use the substitution
u=e^x to make the problem simpler and make it into a quadratic equation.