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Yesterday, my uncle asked me this question:

Prove that $x = 2$ is the unique solution to $3^x + 4^x = 5^x$ where $x \in \mathbb{R}$.

How can we do this? Note that this is not a diophantine equation since $x \in \mathbb{R}$ if you are thinking about Fermat's Last Theorem.

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  • $\begingroup$ @AdiDani: Can you rephrase that please? $\endgroup$
    – P.K.
    Commented Jan 7, 2013 at 13:15
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    $\begingroup$ @DumbCow I've just suggested an edit that does this. Typically $n$ is reserved for use representing natural numbers, and $x$ represents reals. $\endgroup$
    – apnorton
    Commented Jan 7, 2013 at 13:17
  • $\begingroup$ @anorton: Yes, I realized that. Thanks. $\endgroup$
    – P.K.
    Commented Jan 7, 2013 at 13:17
  • $\begingroup$ $$\forall α \in \mathbb{R}, \ α^2 + 4(α + 1) = (α + 2)^2$$ $\endgroup$
    – Mr Pie
    Commented Oct 21, 2017 at 23:15

4 Answers 4

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$$f(x) = \left(\dfrac{3}{5}\right)^x + \left(\dfrac{4}{5}\right)^x -1$$

$$f^ \prime(x) < 0\;\forall x \in \mathbb R\tag{1}$$

$f(2) =0$. If there are two zeros of $f(x)$, then by Rolle's theorem $f^\prime(x)$ will have a zero which is a contradiction to $(1)$.

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  • $\begingroup$ isn't Rolle's Theorem much stronger than what we need $\endgroup$
    – cderwin
    Commented Jan 7, 2013 at 13:21
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For all $x_j>x_i$ and $0<a<1$, $a^{x_i}>a^{x_j}$ .

Hence \begin{align} \left(\frac{3}{5}\right)^{x} + \left(\frac{4}{5}\right)^{x} - 1 < \left(\frac{3}{5}\right)^{2} + \left(\frac{4}{5}\right)^{2} - 1 = 0 \end{align}

for all $x>2$. Hence, there is no solution for $x>2$.

Similarly \begin{align} \left(\frac{3}{5}\right)^{x} + \left(\frac{4}{5}\right)^{x} - 1 > \left(\frac{3}{5}\right)^{2} + \left(\frac{4}{5}\right)^{2} - 1 =0 \end{align}

for all $x<2$.

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Let $f(x)=5^x-4^x-3^x$. Then $f(2)=0$.

If $k>0$, then

$f(2+k)=f(2+k)-f(2)$

$=25(5^k-1)-16(4^k-1)-9(3^k-1)$

$>25(5^k-1)-16(5^k-1)-9(5^k-1)=0$.

If $k<0$, then

$f(2+k)=f(2+k)-f(2)$

$=25(5^k-1)-16(4^k-1)-9(3^k-1)$

$<25(5^k-1)-16(5^k-1)-9(5^k-1)=0$.

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One looks for roots of the function $f:x\mapsto a^x+1-b^x$ with $a=\frac34$ and $b=\frac54$.

  • Since $a\lt1$, the function $x\mapsto a^x$ is decreasing.
  • Since $b\gt1$, the function $x\mapsto b^x$ is increasing.
  • Hence the function $f$ is decreasing.
  • And $f(\pm\infty)=\mp\infty$.

As such, the function $f$ has exactly one root. Since $f(0)=1$ this root is positive.

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