## College Algebra (11th Edition)

Subtracting $2$ to both sides of the equation, the given equation, $\left(\dfrac{1}{2}\right)^x+2=0 ,$ is equivalent to \begin{array}{l}\require{cancel} \left(\dfrac{1}{2}\right)^x=-2 .\end{array} The left side of the equation above is always a positive number for any value of $x$. This will never be equal to the negative value at the right. Hence, this equation does not have a solution.