Answer to Question #152468 in Calculus for Jessie

Question

Find x. e^π = e^(2x) sinx

Answer 1

Here given that ,$e^\pi=e^{2x}.sinx$

$\implies e^{\pi-2x}=sinx$ ......(1)

Again we know that the value of $sinx$ lies between $(-1)$ and $1$ .

Therefore, $-1\leq sinx\leq1$

$\implies-1\leq e^{\pi-2x}\leq1$ [from (1)]

Because $-1\leq e^{\pi-2x}$ is always true, the case $e^{π-2x} \leq 1$ will be considered.

$e^{\pi-2x}\leq1$

$\implies$ $e^{\pi-2x}\leq e^0$

$\implies(\pi-2x)\leq 0$ [As $e^{\pi-2x}$ is a decreasing function]

$\implies x\geq \frac{\pi}{2}$ .......(2)

Because $e^{\pi-2x}\geq 0$ , $\sin(x) \geq0,$ then $0 \leq x \leq \pi.$

But here $x<\frac{\pi}{2}$ does not satisfy the inequality (2).

Therefore $x=\frac{\pi}{2}$ is the required solution of the given equation.