I need to find values of $a$ for the following equation to have two real solutions.
$$a^x=x+2$$
- $(1,\infty)$
- $(0,1)$
- $1/e,e$
- $(1/(e^e), e^e)$
- $(e^{1/e}, \infty)$
This is how I solved this exercise, but I don't understand some things.
I would like to know if there's another way to solve this kind of exercise. I would be happy if I would get some ideas.
Also, from my solution, I don't understand why from that table results just one solution and from the graphic results two solutions. Usually, to see the number of solutions I use this kind of table.
For $a>1$, $f$ decreases from infinity to -1, then increase from -1 to infinity. I'm really confused. Need some suggestions here.
Thank you!
Best Answer
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 $g(b)<0$.