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I don't know where to ask, but I'm trying. I just thing we cannot do hocus pocus methods.

Solve $3^x+4^x=5^x$

Okay, so my friend gave me this equation, and his solution. But I don't belive it holds. It comes here:

"Solution:

$3^x+4^x=5^x\Leftrightarrow \frac{3^x}{3^x}+\frac{4^x}{3^x}=\frac{5^x}{3^x}\Leftrightarrow 1+\left(\frac{4}{3}\right )^x=\left(\frac{5}{3}\right )^x\Leftrightarrow 1+\left(\frac{4}{3}\right )^{\frac{x}{2}\cdot 2}=\left(\frac{5}{3}\right )^{\frac{x}{2}\cdot 2}\Leftrightarrow \left(\frac{4}{3}\right )^{\frac{x}{2}\cdot 2}-\left(\frac{5}{3}\right )^{\frac{x}{2}\cdot 2}=-1\Leftrightarrow \left(\left(\frac{4}{3}\right)^{\frac{x}{2}}\right )^2-\left(\left(\frac{5}{3}\right )^{\frac{x}{2}}\right)^2=-1$

Let $a=\left(\frac{4}{3}\right )^{\frac{x}{2}}$ and let $b=\left(\frac{5}{3}\right )^{\frac{x}{2}}$ then

$a^2-b^2=\frac{-1}{3}\cdot 3 \Leftrightarrow (a-b)(a+b)=\frac{-1}{3}\cdot 3$

Now $a-b=\frac{-1}{3}$ and $a+b=3$ solving the system of equations, we get $a=\frac{4}{3}$ and $b=\frac{5}{3}$ hence we put back.

$a=\left(\frac{4}{3}\right )^{\frac{x}{2}} \Rightarrow \frac{4}{3}=\left(\frac{4}{3}\right )^{\frac{x}{2}} $ and $b=\left(\frac{5}{3}\right )^{\frac{x}{2}} \Rightarrow \frac{5}{3}=\left(\frac{5}{3}\right )^{\frac{x}{2}} $ we see that $x=2$ in both cases, which satisfies the equation."

Amir Hassan
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    How are we meant to get from $a^2-b^2=-1$ to $a-b=\frac {-1}3,,a+b=3$? That seems entirely arbitrary. For example, $a=0,b=1$ satisfies the first equation but not the others. – lulu Jul 03 '18 at 17:43
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    $a^2-b^2=-\frac{1}{4}\cdot{4}$. Why not $a-b=-\frac{1}{4}$ and $a+b=4$ then? – Math Lover Jul 03 '18 at 17:44
  • @MathLover I belive he did multiply with 3 because in the beginning, he divided with $3^x$ on both sides. – Amir Hassan Jul 03 '18 at 17:47
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    https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem – Mason Jul 03 '18 at 17:50
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    It is true that $a-b=-\frac13$ and $a+b=3$ lead to a solution $x=2$. But that makes all the calculations not an iota better than just plugging in $x=2$ into the original equation and seeing that it works. – Hagen von Eitzen Jul 03 '18 at 17:55
  • Analytic solutions are your best bet here. Sticking to , let $f(x)=\left(\frac 54\right)^x-1-\left(\frac 34\right)^x$. Note that $f(0)<0$ and $f(x)$ increases monotonically to $\infty$ as $x$ grows. Thus there is a unique (positive) solution. – lulu Jul 03 '18 at 17:58
  • 3, 4, 5 is a Pythagorean triple so $x=2$ is an obvious solution. We also know that there are no solution among integers greater than 2. – Vasili Jul 03 '18 at 19:10
  • This has been asked many times over, see 1, 2, 3, 4 for example. – dxiv Jul 03 '18 at 22:45

3 Answers3

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It is obvious that $x=2$ is a solution.

To see that there cannot be other (real) solutions: We're looking for solutions of $$ 5^x - 4^x - 3^x = 0 $$ Divide this by $4^x$ (which is always positive) on both sides, and we get $$ (\tfrac54)^x - 1 - (\tfrac34)^x = 0 $$ Here, both the terms $(\tfrac54)^x$ and $-(\tfrac34)^x$ are strictly increasing and the middle $-1$ is a constant. So the entire left-hand side is strictly increasing on all of $\mathbb R$, and therefore has at most one zero -- which must be the one we already know at $x=2$.

hmakholm left over Monica
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If you're asking is this the only possible solution, the logical breakdown I see is at the step where he insists on the particular breakdown of -1 into that product. As far as I can see, you could do the same with any other two factors of -1. Not sure if that will always end up with x=2 or not, but it's an area to investigate.

Alan
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  • But the solution will hold in this case? I solved it numerically and end up with the same answer, my CAS program Maple can't solve it algebraically, only numerically. – Amir Hassan Jul 03 '18 at 17:47
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    The "answer" holds, of course. We can check by hand that substituting $x=2$ will result in a true statement. However, the "solution" -- by which I mean the method of solving the problem -- does not hold, as it posits $a\cdot b = c\cdot d \Rightarrow a=c \wedge b = d$, which is a false statement. If it were true, then it would imply a unique, order-dependent factorization for every number. – Alex Jones Jul 03 '18 at 17:58
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Another method..........

$f(x) = 3^x + 4^x - 5^x$

$f'(x) = \ln(3)3^x + \ln(4)4^x - \ln(5)5^x$

$x_0 = 3$

$x_1 = 3 - \frac{27 + 64 -125}{-82.79} = 2.59$

$x_2 = 2.59 - \frac{-11.15}{-40.73} = 2.32$

$x_3 = 2.32 - \frac{-4.12}{-18.72} = 2.10$

$x_4 = 2.10 - \frac{-.94}{-10.75} = 2.01$

$x_5 = 2.01 - \frac{-.08}{-8.40} = 2.00$

Phil H
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