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Question Number 110471 by abdullahquwatan last updated on 29/Aug/20

the sum of all x values (x^2 −6x+9)^((x−3)/(x−1)) = (x−3)^((x−4))

$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{x}\:\mathrm{values} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}\right)^{\frac{{x}−\mathrm{3}}{{x}−\mathrm{1}}} \:=\:\left({x}−\mathrm{3}\right)^{\left({x}−\mathrm{4}\right)} \\ $$

Answered by Rasheed.Sindhi last updated on 29/Aug/20

(x^2 −6x+9)^((x−3)/(x−1)) = (x−3)^((x−4)) ( (x−3)^2 )^((x−3)/(x−1)) = (x−3)^((x−4)) ^(• ) ((2(x−3))/(x−1))=x−4 2x−6=x^2 −5x+4 x^2 −7x+10=0 (x−2)(x−5)=0 x=2 ∨ x=5 ^• x−3=0,1⇒x=3 ∨x=4 But x=3 produces indeterminate form 0^0 So Solution Set is {2,4,5}

$$\left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}\right)^{\frac{{x}−\mathrm{3}}{{x}−\mathrm{1}}} \:=\:\left({x}−\mathrm{3}\right)^{\left({x}−\mathrm{4}\right)} \\ $$$$\left(\:\left({x}−\mathrm{3}\right)^{\mathrm{2}} \:\right)^{\frac{{x}−\mathrm{3}}{{x}−\mathrm{1}}} \:=\:\left({x}−\mathrm{3}\right)^{\left({x}−\mathrm{4}\right)} \\ $$$$\:^{\bullet\:} \frac{\mathrm{2}\left({x}−\mathrm{3}\right)}{{x}−\mathrm{1}}={x}−\mathrm{4} \\ $$$$\:\:\:\:\:\mathrm{2}{x}−\mathrm{6}={x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4} \\ $$$$\:\:\:\:{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{10}=\mathrm{0} \\ $$$$\:\:\:\left({x}−\mathrm{2}\right)\left({x}−\mathrm{5}\right)=\mathrm{0} \\ $$$$\:\:\:\:{x}=\mathrm{2}\:\vee\:{x}=\mathrm{5} \\ $$$$\:^{\bullet} {x}−\mathrm{3}=\mathrm{0},\mathrm{1}\Rightarrow{x}=\mathrm{3}\:\vee{x}=\mathrm{4} \\ $$$$\:\:{But}\:{x}=\mathrm{3}\:{produces}\:{indeterminate} \\ $$$${form}\:\mathrm{0}^{\mathrm{0}} \\ $$$${So}\:{Solution}\:{Set}\:{is} \\ $$$$\:\:\:\left\{\mathrm{2},\mathrm{4},\mathrm{5}\right\} \\ $$

Commented by abdullahquwatan last updated on 29/Aug/20

3 included?

$$\mathrm{3}\:\mathrm{included}? \\ $$

Commented by Rasheed.Sindhi last updated on 29/Aug/20

Sorry it′s excluded because x=3 gives indeterminate form 0^0

$${Sorry}\:{it}'{s}\:{excluded}\:{because}\:{x}=\mathrm{3} \\ $$$${gives}\:{indeterminate}\:{form}\:\mathrm{0}^{\mathrm{0}} \\ $$

Commented by Rasheed.Sindhi last updated on 29/Aug/20

Thanks abdullah for mentkoning error.

$$\mathcal{T}{hanks}\:{abdullah}\:{for} \\ $$$${mentkoning}\:{error}. \\ $$

Commented by abdullahquwatan last updated on 29/Aug/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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