For which $a$ does the equation $a^x=x+2$ have two solutions?

Your attempt is wrong, sorry: you cannot just use particular cases. And the case $a=-e$ is impossible, because $a^x$ is only defined for $a>0$. The answer should be in terms of $a$, and using a single value is not enough.

Consider the function $f(x)=a^x-x-2$. Then $$ f'(x)=a^x\log a-1 $$ (with $\log$ being the natural logarithm). This doesn't vanish for $0<a\le 1$, so the function can have two zeros only for $a>1$.

In this case the point of minimum is at $$ x=-\frac{\log\log a}{\log a} $$ Set $b=\log a$, for simplicity. Then $a=e^b$ and $a^x=e^{bx}$; we want to evaluate $$ f\left(-\frac{\log b}{b}\right)=e^{-\log b}+\frac{\log b}{b}-2=\frac{-1+\log b-2b}{b} $$ Consider $g(t)=-1+\log t-2t$, for $t>0$; then $g'(t)=\frac{1}{t}-2$, which vanishes for $t=1/2$; since $$ g(1/2)=-1-\log2-1<0 $$ you have the desired answer, because this implies $g(b)<0$.

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Functions

Algebra Precalculus

Exponential Function

Transcendental Equations

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