Finding all positive $a$ such that $a^x=2^x+1$ has only one real solution Find all values of $a \in (0,\infty)$ such that 
$$a^x=2^x+1$$ has only one real solution.
I tried using derivatives, but I couldn't find out. Can somebody help me, please?
 A: Approach 1:
$a^x = 2^x + 1 $, taking ln() (since both sides (functions) are injective)
$x\ln(a) = \ln(2^x+1)$. 
Right now, we gonna totally forget about $x=0$ because clearly for x=0, we have no solutions.
So, let's divide by x:
$\ln(a) = \frac{\ln(2^x+1)}{x}$, and lastly taking $\exp()$ (again both are injective), we get:
$$ a = \exp(\frac{\ln(2^x+1)}{x}) $$
Good, finally something doable. All we need to do now, is to determine whether and where the function $\exp(\frac{\ln(2^x+1)}{x})$ is incjective (not only domain is in our interest but also (and most importantly) the range).
We know the behavior of function $\exp()$. It's increasing and takes every value $y \in(0,+\infty)$.
So let's focus on the inner function : f(x) = $\frac{\ln(2^x+1)}{x}$, $x \in (-\infty,0) \cup (0,\infty) $
$f'(x) = \frac{2^x\ln(2)}{2^x+1}\frac{1}{x} - \frac{1}{x^2}\ln(2^x+1) = \frac{2^x\ln(2)x}{(2^x+1)x^2} - \frac{(2^x+1)\ln(2^x+1)}{x^2(2^x+1)}= \frac{2^x(x\ln(2) - \ln(2^x+1)) - \ln(2^x+1)}{x^2(2^x+1)}$
The denominator is always positive, so we are interested in the sign of:
$g(x) = 2^xx\ln(2) - (2^x+1)\ln(2^x+1)$
As $x\to 0$, $g(x) \to -2\ln(2)$, no matter from which side we approach 0.
As $x \to -\infty$, our $g(x)$ clearly tends to 0 ( $- (2^x+1)\ln(2^x+1)$ part is easy to establish that it tends to 0, and if I were to say something about $2^xx\ln(2)$ limit, there is a known fact, that exponential function ( $2^x$ is indeed an exponential function times some constant) tends "faster" than linear (in fact any polynomial W(x) ))
It would be sufficient for us, to show that $g(x)$ is descreasing (because if it is so, then $g$ would always be less than 0, so $f'$, too, and f would be descreasing and that is something we wanna achieve right now.
$g'(x) = 2^x\ln(2) + x\ln^2(2)2^x - 2^x\ln(2)(\ln(2^x+1)) - \frac{(2^x+1)}{(2^x+1)}2^x\ln(2) = \\ = 2^x\ln(2)[1+x\ln(2)-\ln(2^x+1) - 1]  = 2^x\ln(2)[x\ln(2) - \ln(2^x+1)]  $
Hosanna! Only important term is that inside the braces (because we know the sign of $2^x\ln(2)$ (+). Looking at:   $x\ln(2) - \ln(2^x+1)$, we can just prove it is always less than 0!!!.
$ x\ln(2) < \ln(2^x+1) $
$ \ln(2^x) < \ln(2^x+1) $
Q.E.D cause $\ln$ is an increasing function!
We're home! That means (if we apply everything to the very beggining, that our $f(x)$ is decreasing, so function $\exp(f(x)) = \exp(\frac{\ln(2^x+1)}{x})$ is decreasing, too! (But importantly on each set $(-\infty,0)$, $(0,+\infty)$ )
That means, we must establish limits, as $\exp(f(x))$ tends to $\{-\infty,0^-,0^+,+\infty\}$
$$ \lim_{x\to -\infty} \exp(f(x)) = 1 $$
$$ \lim_{x\to 0^-} \exp(f(x)) = 0 $$
$$ \lim_{x\to 0^+} \exp(f(x)) = +\infty $$
$$ \lim_{x\to +\infty} \exp(f(x)) = 2 $$
And now, we're only reading what we've just achieved:
If we're looking at $x\in(-\infty,0)$, our function decreases from 1 to 0, so $a\in(0,1)$ do the job. Taking a look at $x \in (0,+\infty)$, our function descreases from $+\infty$ to 2, so again, $a \in(2,+\infty)$ do the job.
Summary: For every $a \in(0,1) \cup (2,+\infty)$ the equation $2^x+1 =a^x$ has only one real solution.
Approach 2:
Instead of treating "a" as a function of x ( that is a(x) ) , we'll now really dig into every side one by one:
Consider 2 functions ( in fact one would be a whole family of functions):
$$f:R\to R, f(x) = 2^x+1$$
$$g_a:R\to R, g_a(x) = a^x ; a\in(0,1) \cup (1,+\infty) $$
(I haven't considered case $a=1$, because it clearly cannot hold, cause $2^x$ cannot be 0)
Function $f$ is clearly increasing (You can always find it derivative, as $f'(x) = 2^x\ln(2)$ which is positive for all $x\in R$).
$\lim_{x \to -\infty} f(x) = 1$
$\lim_{x \to +\infty} f(x) = +\infty $
We'll need those limits later on.
Now, take a look at function $g_a$ and find it derivative:
$g_a'(x) = a^x\ln(a)$
Now, we cannot say it is positive/negative for all $x\in R$ cause it depends on "a".
Consider 2 cases:
Case1:  $(a \in (0,1))$ 
$g_a'$ is negative, so $g_a(x)$ is decreasing. Finding $\lim_{x\to -\infty} g_a(x) = +\infty$, and $\lim_{x \to +\infty} g_a(x) = 0$, We clearly see, function $g_a$ DO intersect with the function $f$. So all $a\in(0,1)$ do the job.
Case2: $(a \in (1,+\infty)) $
$g_a'$ is positive, so $g_a$ is increasing. Hovewer, this case isn't as easy as the previous one, but still we can deal with it. Again, we need limits.
$\lim_{x\to -\infty} g_a(x) = 0$
$\lim_{x\to +\infty} g_a(x) = +\infty$
So unfortunatelly its behaviour is pretty similar to function f, with one exception: f "starts" from 1.
For $g_a$ to intersect $f$, it is needed and sufficient that:
$$ \forall_{x \in R} \ : \ \ g_a'(x) > f'(x) $$
One may ask why? Because we need $g_a$ to intersect $f$ at only one point (this is very important). We know that from the beggining $g_a$ is less than $f$ (by the beggining I mean $\lim_{x\to -\infty}$). So if we want $g_a$ to intersect $f$, it growing ratio must be "bigger" and it must STAY bigger, because we do not want $f$ to "outrun" $g_a$. Hope it is clear.
$ g_a'(x) = a^x\ln(a) $, while $f'(x) = 2^x\ln(2) $
Since both $\log$ and $\exp$ functions are increasing, it is really easy to establish, that if $a=2$ we have equality, and our interesting case is if and only if $a\in(2,+\infty)$
Summary: Taking both together, we again ( Hurra! ) get $a \in (0,1) \cup (2,+\infty)$ are the "good" ones.
Hope you'll find it helpful.
A: Hint:
Plot the functions $a^x$ and $2^x+1$ for the following values of $a$:


*

*$a=\frac12$

*$a=\frac13$

*$a=\frac14$

*$a=1$

*$a=1.1$

*$a=1.2$

*$a=1.3$

*$a=2$

*$a=3$

*$a=4$
Can you notice a pattern?
A: $$a^x=2^x+1$$
$$a^x-2^x=1$$
Let's say $a^x-2^x\approx(a-2)^x$ , hence:


*

*$a\neq2$ because $0^x$ 

*for $a\in(0;2)$ we get $\alpha^x=1$ where $\alpha<0\implies x=0$ 

*$2<<a:$
$$(a-2)^x=1$$
$$x=0=log_{a-2}1$$

*For small $a>2$ -a numerical method like Dr. Sonnhard Graubner said

