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

[solakis1]

Solve the following equation:
$(x^2-7x+11)^{x^2-13x+42}=1$

 

[topsquark]

[sp]
[math](x^2 - 7x + 11)^(x^2 - 13x + 42) = 1[/math]
[math](x^2 - 13x + 42) ln(x^2 - 7x + 11) = 0[/math]

Thus
[math]x^2 - 13x + 42 = 0 \implies x = \{ 6, ~ 7 \}[/math]
[/sp]

-Dan

 

[solakis1]

No

 

[topsquark]

[sp]
Oh, of course. Too tired.
[math]ln(x^2 - 7x + 11) = 0 \implies x^2 - 7x + 11 = 1 \implies x = \{ 2, ~ 5 \}[/math]
as well.
[/sp]

-Dan

 

[kaliprasad]

Solve the following equation:
$(x^2-7x+11)^{x^2-13x+42}=1$

Spoiler: my solution

There are 3 cases
1) exponent is 0
this gives $x^2-13x + 42= 0$ giving x = 6 or 7
2) base is 1
$x^2-7x+11=1$ or $x^2-7x + 10 = 0$ giving x = 2 or 5
3) base is -1 and exponent is even
giving
$x^2 - 7x + 11= -1$ or $x^2-7x + 12 = 0$ giving x = 3 or 4
for both cases exponenent = x(x-13) + 42 even

combinng all 3 above
We get $ x \in \{ 2,3,4,5,6,7\}$

 

[solakis1]

very good

 

[SatyaDas]

Spoiler: my solution

There are 3 cases
1) exponent is 0
this gives $x^2-13x + 42= 0$ giving x = 6 or 7
2) base is 1
$x^2-7x+11=1$ or $x^2-7x + 10 = 0$ giving x = 2 or 5
3) base is -1 and exponent is even
giving
$x^2 - 7x + 11= -1$ or $x^2-7x + 12 = 0$ giving x = 3 or 4
for both cases exponenent = x(x-13) + 42 even

combinng all 3 above
We get $ x \in \{ 2,3,4,5,6,7\}$

I wonder about existence of complex solutions. Can't it happen that x is a complex number such that $(x^2-7x + 11)^n = 1$ and $x^2-13x + 42= n$ where $n\in Z^+$? It appears there are other alternatives as well if we allow complex solutions.

 

[SatyaDas]

Solve the following equation:
$(x^2-7x+11)^{x^2-13x+42}=1$

Real solutions have already been given so I am focusing on complex solutions.
To find solutions in complex plane we can rewrite the equation as $(x^2-7x+11)^{x^2-13x+42}=\exp \left({2k\pi i}\right)$ where $k \in Z$.
Now, we can rearrange the terms to write
$$
x=\frac{\exp \left(\frac{2k \pi i}{x^2-13 x+42}\right)-11}{x-7}.
$$
This form gives us the opportunity to find the solution iteratively. We can start with any integer value of k and starting guess for x, e.g. $x=0$ and iteratively compute next approximation of x. For example here are some complex solutions found numerically:

k	x
0	2
1	2.00793 - 0.105303 I
2	2.03253 - 0.214414 I
3	2.07647 - 0.333058 I

There is probably many solutions to the equation.