Question

Find the roots of f(x)=(e^x-e^π)(e^x-π) where e denotes Euler's number

Answer 1

The given equation is $f(x)=(e^x-e^{\pi})(e^x-\pi)=0$ . This equation is a quadratic equation in $y=e^x$. The solutions of the quadratic are

[tex](y-e^{\pi})(y-\pi)=0\\ y=e^{\pi},\pi[/tex]

Hence, when $y=e^{\pi}$, $e^x=e^{\pi},x=\pi$\\

when $y=\pi$, [$e^x=\pi,x=\ln(\pi)$.

Thus the roots of the given equation are $x=\ln(\pi),\pi$.

Answer 2

Answer:

The roots are x =$\pi$ and x = $log(\pi)$

Step-by-step explanation:

f(x) = $(e^{x}-e^{\pi})(e^{x} -\pi)$ where e is the Euler number.

To find the roots, put f(x) = 0

$(e^{x}-e^{\pi})(e^{x} -\pi)$ = 0

$(e^{x}-e^{\pi})$ = 0

$e^{x} = e^{\pi}$

$x = \pi$

$(e^{x} -\pi)$ = 0

$e^{x} = \pi$

$x = log(\pi)$

Therefore, the roots are$x = \pi and x = log(\pi)$