Question

Find all the real solutions to the exponential equation
$e^{2x}-3e^x+2=0$

• A. $1,2$
• B. $e^2$
• C. no real solutions
• D. $0,\ln \left(2\right)$
• E. $e^3$

Answer 1

The correct option is D $0,\ln \left(2\right)$

Consider, $e^{2}x-3e^x+2=0$

Let $z=e^x$......................(1)

$z^2-3z+2=0$

Finding the roots of the above quadratic equation, we get

$z^2-3z+2=0$

$z^2-2z-z+2=0$
$z(z-2)-1(z-2)=0$
$z=1$ and $z=2$

Substituting the value of $z=1$ in equation (1)

$1=e^x$

$\ln \left(1\right)=\ln \left(e^x\right)$

$x=\ln \left(1\right)$

$x=0$ .... (as $\ln \left(1\right)=0$)

Now substituting the value of $z=2$ in equation (1)

$2=e^x$

$\ln \left(2\right)=\ln \left(e^x\right)$

$x=\ln \left(2\right)$

Therefore, $x=\ln \left(2\right)$ and $x=0$.