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Is there a way to solve $5^x-4^x-3^x-2^x-26=0$ by hand?

Added for clarity:

I can test values and quickly find $x=3$ is a solution and can show that it is the only one.

What I am curious about is if there is some nice more general method to arrive at this solution directly.

I tried messing around with logs a bit and ended up with something that might have been better, but I couldn't see a way forward:

$y^{\ln{5}}-y^{\ln{4}}-y^{\ln{3}}-y^{\ln{2}}-26=0$

David Taub
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    Try integer values starting from 0 and you will get the answer in a few guesses. Then all you need to do is prove this is the only solution. – JC12 Aug 07 '20 at 11:17
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    With one solution given by $x=3$, the obvious move is to examine $f'(x).$ If you can show that it is monotonic then $x=3$ must be a unique solution. – user2661923 Aug 07 '20 at 11:18
  • See https://math.stackexchange.com/questions/3780945/if-left-sqrt2-rightx-left-sqrt3-rightx-left-sqrt13-right-fra – lab bhattacharjee Aug 07 '20 at 11:19
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    The hidden question is being begged: what is the context of the problem; why was the OP confronted with this question at this time? Are there theorems (or similar but easier problems) that the OP was recently confronted with that are pertinent. Off-topic: what if the last term had been -25 instead of -26. In that case, is there a closed form expression for $x$? – user2661923 Aug 07 '20 at 11:22
  • I probably should have been more clear - I am aware of $x=3$ being the only solution and can show that it is once found, but I was wondering if there was a nice way to find the solution if it wasn't easy to just test to it. For example, if it were irrational instead or just harder to find. – David Taub Aug 07 '20 at 11:32
  • @Kevin: if the root at $3$ is known, what is the point computing it by Newton-Raphson ? –  Aug 07 '20 at 12:17

3 Answers3

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If $26$ is changed to any number, inspection would just give bounds for the solution and you would need numerical methods.

Consider that you look for the zero of function $$f(x)=5^x-4^x-3^x-2^x-k$$ which is not very nice (just plot it). At the opposite $$g(x)=x \log(5)-\log(4^x+3^x+2^x+k)$$ is close to the problem of the intersection of two straight lines.

Make a Taylor expansion around $x=1$ to make $$g(x)=(\log (5)-\log (k+9))+$$ $$\frac{ (k \log (5)+9 \log (5)-4 \log (4)-3 \log (3)-2 \log (2))}{k+9}(x-1)+O\left((x-1)^2\right)$$ and solve for $x$. $$x=\frac{(k+9) \log (k+9)-4 \log (4)-3 \log (3)-2 \log (2)}{(k+9) \log (5)-4 \log (4)-3 \log (3)-2 \log (2)}$$

For $k=26$, this gives $x \sim 2.477$. Notice that this is exactly the first iteration of Newton method.

Doing the same with Halley's method, the estimate would be $2.735$ and with Householder method $2.926$.

Edit

As @Oscar Lanzi answered, we must have $5^x > k$. So, let $$x_0=\frac{\log (k)}{\log (5)}$$ and make a Taylor expansion of first order around this point (or, equivalently, one single iteration of Newton method). This would give as a better estimate $$x_1={x_0}-\frac{{x_0} \log (5)-\log \left(2^{x_0}+3^{x_0}+4^{x_0}+k\right)}{\log (5)-\frac{2^{x_0} \log (2)+3^{x_0} \log (3)+4^{x_0} \log (4)}{2^{x_0}+3^{x_0}+4^{x_0}+k}}$$ Trying for $k=10^p$, the results would be $$\left( \begin{array}{cccc} p & x_0 & x_1 & \text{exact} \\ 1 & 1.43068 & 2.42347 & 2.72473 \\ 2 & 2.86135 & 3.44185 & 3.52993 \\ 3 & 4.29203 & 4.64758 & 4.67284 \\ 4 & 5.72271 & 5.95128 & 5.95909 \\ 5 & 7.15338 & 7.30590 & 7.30849 \\ 6 & 8.58406 & 8.68855 & 8.68944 \\ 7 & 10.0147 & 10.0876 & 10.0880 \\ 8 & 11.4454 & 11.4969 & 11.4970 \\ 9 & 12.8761 & 12.9128 & 12.9128 \\ 10 & 14.3068 & 14.3331 & 14.3331 \end{array} \right)$$

Claude Leibovici
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  • For a given $k$ we can use the bounding conditions to generate a better initial value for $x$, which then enhances the convergence of the Newton step. Here we know $2<x<5$ and we can choose $x=3$ or $x=4$ as a Taylor expansion point. If we choose the former to keep numbers in the calculation smaller, well ... . – Oscar Lanzi Aug 07 '20 at 13:30
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    @OscarLanzi. For sure ! We can play a lot with inequalities and have very nice estimates. Have a nice day. Cheers :-) – Claude Leibovici Aug 07 '20 at 13:46
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We cannot solve this equation analytically without assuming an integer solution, but we can show that if the equation has an integer solution it must be $x=3$.

First off $5^x>26$ forcing $x>2$. On the other hand, $5^x$ cannot be allowed to absolutely dominate $4^x+3^x+2^x+26<3(4^x)+26$, so $x<5$ ($5^5=3125, 3×4^5+26=3098$). Finally, if $x$ is even the equation fails $\bmod 4$ because $5^x-3^x$ (a difference of odd squares), $4^x$ and $2^x$ all have residue zero.

So any integer candidate must be odd and satisfy $2<x<5$, rendering $x=3$ the sole survivor.

Oscar Lanzi
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For a resolution "by hand", you have few options. You can try a graphical method and evaluate the function at a few integer points (the computation is easy). In this case a rough sketch should be enough to see that there is a single root near $3$.