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Question Number 68935 by Maclaurin Stickker last updated on 17/Sep/19

Find all values for x:  (x^2 −7x+11)^(x^2 −13x+42) =1  (Easy)

$${Find}\:{all}\:{values}\:{for}\:\boldsymbol{{x}}: \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{11}\right)^{{x}^{\mathrm{2}} −\mathrm{13}{x}+\mathrm{42}} =\mathrm{1} \\ $$$$\left({Easy}\right) \\ $$

Answered by $@ty@m123 last updated on 17/Sep/19

x=7,6  x=2,5  (Calculated in mind.)

$${x}=\mathrm{7},\mathrm{6} \\ $$$${x}=\mathrm{2},\mathrm{5} \\ $$$$\left({Calculated}\:{in}\:{mind}.\right) \\ $$

Answered by Rasheed.Sindhi last updated on 17/Sep/19

Either base=1 Or exponent=0  (x^2 −7x+11)^(x^2 −13x+42) =1  x^2 −7x+11=1 ∣  x^2 −13x+42=0  x^2 −7x+10=0 ∣ (x−6)(x−7)=0  (x−2)(x−5)=0 ∣ x=6,7  x=2,5

$$\mathcal{E}{ither}\:{base}=\mathrm{1}\:\mathcal{O}{r}\:{exponent}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{11}\right)^{{x}^{\mathrm{2}} −\mathrm{13}{x}+\mathrm{42}} =\mathrm{1} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{7x}+\mathrm{11}=\mathrm{1}\:\mid\:\:{x}^{\mathrm{2}} −\mathrm{13}{x}+\mathrm{42}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{7x}+\mathrm{10}=\mathrm{0}\:\mid\:\left(\mathrm{x}−\mathrm{6}\right)\left(\mathrm{x}−\mathrm{7}\right)=\mathrm{0} \\ $$$$\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}−\mathrm{5}\right)=\mathrm{0}\:\mid\:\mathrm{x}=\mathrm{6},\mathrm{7} \\ $$$$\mathrm{x}=\mathrm{2},\mathrm{5} \\ $$$$ \\ $$

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