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?

  • 2
    $\begingroup$ You will need a numerical method $\endgroup$ Oct 15, 2018 at 10:19
  • 1
    $\begingroup$ This will depend on $\;a\;$ (what is it?), and most probably you'll need, as Sonnhard says, a numerical method. $\endgroup$
    – DonAntonio
    Oct 15, 2018 at 10:21
  • 2
    $\begingroup$ 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 $x$? $\endgroup$
    – Blue
    Oct 15, 2018 at 10:24
  • 1
    $\begingroup$ 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? $\endgroup$
    – lulu
    Oct 15, 2018 at 10:26
  • 1
    $\begingroup$ 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. $\endgroup$ Oct 15, 2018 at 10:30

3 Answers 3


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.

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


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?

  • $\begingroup$ Can I have a proof?In an exam you don't have time for plotting. $\endgroup$
    – tyuiop
    Oct 15, 2018 at 10:38
  • 6
    $\begingroup$ @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.. $\endgroup$
    – 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 $\alpha<0\implies x=0$
  • $2<<a:$ $$(a-2)^x=1$$ $$x=0=log_{a-2}1$$
  • For small $a>2$ -a numerical method like Dr. Sonnhard Graubner said
  • $\begingroup$ It's said to be (0,1] and $e^2$ but I don't know why. $\endgroup$
    – tyuiop
    Oct 15, 2018 at 12:36
  • $\begingroup$ @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 $g(a;x)=a^x$ $\endgroup$ Oct 15, 2018 at 13:10
  • $\begingroup$ @AndreiGabor Please include the letter $a$ or $x$ to avoid confusion $\endgroup$ Oct 15, 2018 at 13:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.