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

• MHB
• solakis1
In summary, the equation $(x^2-7x+11)^{x^2-13x+42}=1$ can be rewritten as $(x^2-7x+11)^{x^2-13x+42}=\exp \left({2k\pi i}\right)$ in order to find complex solutions. These solutions can be found iteratively by starting with a value of k and a starting guess for x. There may be many solutions to this equation.
solakis1
Solve the following equation:
$(x^2-7x+11)^{x^2-13x+42}=1$

[sp]
$$\displaystyle (x^2 - 7x + 11)^(x^2 - 13x + 42) = 1$$
$$\displaystyle (x^2 - 13x + 42) ln(x^2 - 7x + 11) = 0$$

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

-Dan

No

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

-Dan

solakis said:
Solve the following equation:
$(x^2-7x+11)^{x^2-13x+42}=1$
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 $x \in \{ 2,3,4,5,6,7\}$

very good

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.

solakis said:
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
$$\displaystyle x=\frac{\exp \left(\frac{2k \pi i}{x^2-13 x+42}\right)-11}{x-7}.$$
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.

## What is the equation asking to be solved?

The equation is asking for the values of x that satisfy the equation $(x^2-7x+11)^{x^2-13x+42}=1$.

## How do I solve this equation?

To solve this equation, you can use logarithms or algebraic manipulation to isolate x and find its possible values. You can also use a graphing calculator to visually see the points of intersection.

## What are the possible values of x?

The equation has an infinite number of possible values for x. However, there are certain values that can be determined through algebraic manipulation or graphing. These values include x = 1, x = 5, and x = 7.

## Can I use any value for x?

No, not all values of x will satisfy the equation. Certain values, such as x = 0, will result in an undefined solution. It is important to check your solutions to ensure they are valid.

## What is the significance of the number 1 in the equation?

The number 1 in the equation is the base of the exponential function. This means that any value raised to the power of 0 will equal 1, resulting in the equation being satisfied. It is also the only value that will result in a solution for x = 0.

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