Question

Find the real solutions, if any, of the equation$e^{x^2-9}=1$

Short answer

Solution set is$(-3,3)$

Step-by-step solution

Step 1. The given information is 

$e^{x^2-9}=1$

Step 2. Using exponent identity e0=1 we get,

$$\begin{gathered} e^{x^2-9}=e^0 \\ {x}^2-9=0 \\ {x}^2-3^2=0 \\ (x+3)(x-3)=0 \\ x+3=0,x-3=0 \\ x=-3,x=3 \end{gathered}$$

solution set is$(-3,3)$