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# Expression problem

Solve for $$x$$:

$\large \displaystyle{(x^2-7x+12)^{x^2-3x+2} = 1}$

Note by Cedie Camomot
1 year, 8 months ago

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Split into three cases

Case 1: $$x^2-7x+12 = 1$$, and $$x^2-3x+2$$ is any real number

The first case will give $$\displaystyle x = \frac{7\pm\sqrt{5}}{2}$$, which, when substitute to $$x^2-3x+2$$, gives the exponent of $$5\pm\sqrt{5}$$, which is valid, as $$1^{5\pm\sqrt{5}} = 1$$, hence one of the answers is $$\displaystyle x = \frac{7\pm\sqrt{5}}{2}$$

Case 2: $$x^2-7x+12$$ is nonzero real number, and $$x^2-3x+2=0$$

This will give $$x = 1, 2$$, and it is valid as $$6^0 = 2^0 = 1$$.

Case 3: $$x^2-7x+12 = -1$$, and $$x^2-3x+2$$ is even numbers

This will give $$\displaystyle x = \frac{7\pm\sqrt{3}i}{2}$$ (where $$i = \sqrt{-1}$$), and the exponents is $$3\pm\sqrt{3}i$$. It will be $$(-1)^{3\pm\sqrt{3}i}$$, which gives a result in transcendental number and never equals 1

Therefore, there are four solutions, which is $$1, 2, \phi+3, 4-\phi$$ (where $$\displaystyle \phi = \frac{1+\sqrt{5}}{2}$$)

- 1 year, 8 months ago

Pretty sure you meant even in the third case.

- 1 year, 8 months ago

Yes, I wrote that incorrectly. Thanks.

- 1 year, 8 months ago

Use factoring algebra. Like this : (x^2 -7x + 12) x^2-3x +2 = 1 (x-3) (x-4) (x-1) (x-2) (x=3 x=4 ) (x=1 x=2) Try to apply one by one the value of x in the algebra problems

- 1 year, 8 months ago

You mean: $$(x^2 -7x + 12) x^2-3x +2 = 1$$ $$(x-3) (x-4) (x-1) (x-2)$$ $$x=3 x=4$$ $$x=1 x=2$$?

- 1 year, 8 months ago

1

- 1 year, 8 months ago

x^2-3x+2=0 => x^2-2x-x+2 =0 x= 2, x=1

- 1 year, 8 months ago