- #1

solakis1

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Solve the following equation:

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

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

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- MHB
- Thread starter solakis1
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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.

- #1

solakis1

- 422

- 0

Solve the following equation:

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

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

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- #2

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\(\displaystyle (x^2 - 7x + 11)^(x^2 - 13x + 42) = 1\)

\(\displaystyle (x^2 - 13x + 42) ln(x^2 - 7x + 11) = 0\)

Thus

\(\displaystyle x^2 - 13x + 42 = 0 \implies x = \{ 6, ~ 7 \}\)

[/sp]

-Dan

- #3

solakis1

- 422

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No

- #4

- 2,018

- 810

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

- #5

kaliprasad

Gold Member

MHB

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solakis said:Solve the following equation:

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

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\}$

- #6

solakis1

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very good

- #7

SatyaDas

- 22

- 0

kaliprasad said:

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.

- #8

SatyaDas

- 22

- 0

Real solutions have already been given so I am focusing on complex solutions.solakis said:Solve the following equation:

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

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 |

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

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.

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.

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.

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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