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?

• You will need a numerical method Oct 15, 2018 at 10:19
• This will depend on $\;a\;$ (what is it?), and most probably you'll need, as Sonnhard says, a numerical method. Oct 15, 2018 at 10:21
• I re-formatted your question slightly, but the meaning isn't entirely clear as it stands. Are you looking for $a$ such that the equation holds for all $x$? are you looking for $a$ such that the equation holds for some $x$?
– Blue
Oct 15, 2018 at 10:24
• Not sure what you are asking. If $a>2$ then there is a solution, if $a≤2$ then there isn't (clearly). Is that what you meant?
– lulu
Oct 15, 2018 at 10:26
• Take $ln$ both sides (natural logarithm) so you can write the left hand side as $xln(a)$ and then divide by $x$ (considering it different from 0), and finally add the eponential both sides to get the value of a. Oct 15, 2018 at 10:30

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.

• (+1) You did so much but no one bothered to upvote ;-( Oct 15, 2018 at 15:51
• I've done it right now, no problem. I've done it to help, not to get any points :P Oct 15, 2018 at 15:53
• Anyways. Points are the way good answers get rewarded on this site and they deserve it as an encouragement for new users Oct 15, 2018 at 15:55

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?

• Can I have a proof?In an exam you don't have time for plotting. Oct 15, 2018 at 10:38
• @AndreiGabor This site isn't here to generate proofs on a whim, but to help you solve problems. My hint should give you a lot of insight into what is happening with the equation $a^x=2^x+1$ for various values of $a$, and should help you get the idea that you should be able to prove. But until you know what you need to prove, trying to prove it is useles..
– 5xum
Oct 15, 2018 at 10:40

$$a^x=2^x+1$$ $$a^x-2^x=1$$ 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-2)^x=1$$ $$x=0=log_{a-2}1$$
• For small $$a>2$$ -a numerical method like Dr. Sonnhard Graubner said
• It's said to be (0,1] and $e^2$ but I don't know why. Oct 15, 2018 at 12:36
• @AndreiGabor (is a matter if you mean $a$ or $x$) draw graphs- this is the geometrical interpretation of your problem: $f(x)=2^x+1$ and some $g(a;x)=a^x$ Oct 15, 2018 at 13:10
• @AndreiGabor Please include the letter $a$ or $x$ to avoid confusion Oct 15, 2018 at 13:15