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Find the solution of ${\left( {{x^2} - 7x + 11} \right)^{\left( {{x^2} - 13x + 42} \right)}} = 1$

My approach is as follow $\phi \left( x \right) = f{\left( x \right)^{g\left( x \right)}}$

Case 1: $f\left( x \right) = 1;x = 2,5$

Case 2: $f\left( x \right) \ne 0;g\left( x \right) = 0;x = 6,7\& f\left( 6 \right) \ne 0;f\left( 7 \right) \ne 0$

Case 3: $f\left( x \right) = - 1;g\left( x \right) =$ even ;$x = 3,4\& f\left( 3 \right) = 12;f\left( 4 \right) = 6$

I have one doubt in Case 3, I am analysing the case $x^x$, in such scenario $x>0$.

Also if $y=a^x$, then if $a=0,x \in R-0$ and $a>0$

$\phi \left( x \right) = f{\left( x \right)^{g\left( x \right)}}$ is also a type of exponential function, they I presume that f(x) is not negative so why they have taken $f(x)=-1$ for case 3, please elaborate

Samar Imam Zaidi
  • 8,800

2 Answers2

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Eric is correct, but you need to exclude the case where $|f(x)|=0$, because $0^{0}$ is indeterminate form. $f(x)$ can be negative, as long as $g(x)$ makes $\phi(x)$ defined and real, in our case as long as $g(x)$ (even).

Mr. Calculus
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$$\left ( x^2-7x+11 \right )^{\left ( x^2-13x+42 \right )}=1\Leftrightarrow \left ( x^2-7x+11 \right )^{\left ( x^2-13x+42 \right )} =\left ( x^2-7x+11 \right )^0$$ $$\left ( x^2-7x+11-1 \right )\left ( x^2-13x+42 \right )=0\Leftrightarrow \left ( x^2-7x+10 \right )\left ( x^2-13x+42 \right )=0$$ $$x^2-7x+10=0\Rightarrow x=\left \{ 2,5 \right \}$$ $$x^2-13x+42=0\Rightarrow x=\left \{ 6,7 \right \}$$ You need to check the roots: $$x^2-7x+11>0$$ All four roots fit!

Fk-91
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