Please help to find out the general solution of the following equation : ${ (x^2 - 7x + 11)}^{x^2 - 13x +42}=1$

Question

For this equation :

${ (x^2 - 7x + 11)}^{x^2 - 13x +42}=1$

The integer solutions of $x$ found by WolframAlpha using inverse (logarithmic) function are $ 2 , 5 , 6 , 7 .$ Why it cannot find the other solutions 3 and 4 ? Is there any full proof method to find all the solutions at once ?

Related: https://math.stackexchange.com/questions/1406796/how-do-you-solve-x2-5x5x2-36-1 — player3236, Dec 21 '20 at 06:52
https://math.stackexchange.com/questions/384090/find-all-real-numbers-x-for-which-frac8x27x12x18x-frac76 — lab bhattacharjee, Dec 21 '20 at 07:04
You cannot find all solutions since there are infinitely many of them. The relation $b^0=1$ is true for infinitely many values of $b.$ — Allawonder, Dec 21 '20 at 07:33

score 1 · Accepted Answer · answered Dec 21 '20 at 07:39

I bet the reason why Wolfram can't find solutions 3 and 4 for the equation

$$ (x^2-7x+11)^{x^2-13x+42}=1 $$

is because it thinks that the domain of the function in LHS is $x^2-7x+11 \gt 0$. And this does make sense. In my school textbook the domain of exponential function $a^x$ is given by $a>0, a \ne 1, x \in \mathbb{R}$. There is no definition to calculate raise of some arbitrary negative base to the power of some arbitrary real number. Your function clearly is not the exponential, since the base is not a constant.

But you can do it manually. Firstly, let's assume that $x^2-7x+11>0$. Then let's take the logarithm from both sides(since it is monotonic, $\ln a= \ln b \iff a=b$): $$ (x^2-13x+42)\ln(x^2-7x+11)=0. $$

The equality holds when $x^2-13x+42=0 \space \text{and in the same time} \space x^2-7x+11>0$ or if simply $x^2-7x+11=1$.

$$ x^2-13x+42=0 \iff x=6 \space \text{or} \space x=7. \\ 6^2-7 \cdot 6+11=36-42+11>0 \\ 7^2-7 \cdot 7+11>0 $$

Then both $x=6$ and $x=7$ satisfy our conditions.

Now consider another case: $$ x^2-7x+11=1 \iff x=2, x=5. $$

We solved our equation assuming the domain is $x^2-7x+11>0$. We can not solve it when $x^2-7x+11 \le 0$ since the function is not well-defined for arbitrary case. But we do know how to raise negative integer to the positive integer power.It means that if we assume that $x^2-7x+11$ is a negative integer, then obviously it should be equal to $(-1)$ and the power the should be positive even integer:

$$ x^2-7x+11=-1 \iff x=3, x=4 \\ \text{then} \space x^2-13x+42 \space \text{should be positive even integer. Let's check it: }\\ 3^2-13 \cdot 3 + 42=9-39+42>0 \\ 4^2-13 \cdot 4 + 42=16-52+42> 0 $$

So, your favourite $x=3$ and $x=4$ also satisfy our equation. We are done now! Any questions are greatly appreciated!

@NilabjaSaha mark this question as an answer to your question if you find it useful. — Levon Minasian, Dec 21 '20 at 07:44

user376343 · Answer 2 · 2020-12-21T07:52:29.900

We want to solve in $\mathbb{Z}$ the equation ${ (x^2 - 7x + 11)}^{x^2 - 13x +42}=1.$

Since $a^b=1$ for some $a,b\in \mathbb{Z}$ only if

$a=1$ or
$a\neq 0$ and $b=0$ or
$a=-1$ and $b$ is even,

we solve for each of listed cases:

${x^2 - 7x + 11}=1 \iff x=2 \quad \text{or}\quad x=5,$
${x^2 - 13x + 42}=0 \iff x=6 \quad \text{or} \quad x=7,$
${x^2 - 7x + 11}=-1 \iff x=3 \quad \text{or} \quad x=4.$

It is easy to check that each of the above numbers satisfies the equation.

REMARK

The "difficulty" of WolframAlpha is related to the use of a function inverse to exponential function. It is not defined if the basis of the exponential is not positive. In the above solution, this corresponds to the case 3.

This question and solutions may also be useful https://math.stackexchange.com/questions/2943102/solution-to-the-equation-of-a-polynomial-raised-to-the-power-of-a-polynomial/2943136#2943136 — user376343, Dec 21 '20 at 07:54

Please help to find out the general solution of the following equation : ${ (x^2 - 7x + 11)}^{x^2 - 13x +42}=1$

2 Answers2