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Question Number 191254 Answers: 0 Comments: 6

a=((x−a)/(x−b)) solve for x did i make anything wrong in the following? starpoint: ln∣a∣=ln∣((x−a)/(x−b))∣ ln∣a∣=ln∣x−a∣−ln∣x−b∣ ln∣a∣=ln∣(x/a)∣−ln∣(x/b)∣ ln∣a∣=ln∣x∣−ln∣a∣−(ln∣x∣−ln∣b∣) ln∣a∣=ln∣x∣−ln∣a∣−ln∣x∣+ln∣b∣ 2∙ln∣a∣=ln∣b∣ e^(ln∣a^2 ∣) =e^(ln∣b∣) a^2 =b if a^2 =b then a=((x−a)/(x−b)) a∙(x−b)=(x−a)⇔x−b≠0 a∙x−a∙b=x−a a∙x−x=a∙b−a x(a−1)=a∙a^2 −a x=((a^3 −a)/(a−1))⇔a−1≠0 x=((a(a^2 −1))/(a−1)) x=((a∙(a−1)∙(a+1))/(a−1)) x=a(a+1) x=b+a answer: x=a^2 +a or x=b+a, and x,a,b ∉ C so now the question is: what if x,a,b ∈ C?

$$ \\ $$$${a}=\frac{{x}−{a}}{{x}−{b}}\:{solve}\:{for}\:{x} \\ $$$$\mathrm{did}\:\mathrm{i}\:\mathrm{make}\:\mathrm{anything}\:\mathrm{wrong}\:\mathrm{in}\:\mathrm{the}\:\mathrm{following}? \\ $$$$ \\ $$$$\mathrm{starpoint}: \\ $$$${ln}\mid{a}\mid={ln}\mid\frac{{x}−{a}}{{x}−{b}}\mid \\ $$$${ln}\mid{a}\mid={ln}\mid{x}−{a}\mid−{ln}\mid{x}−{b}\mid \\ $$$${ln}\mid{a}\mid={ln}\mid\frac{{x}}{{a}}\mid−{ln}\mid\frac{{x}}{{b}}\mid \\ $$$${ln}\mid{a}\mid={ln}\mid{x}\mid−{ln}\mid{a}\mid−\left({ln}\mid{x}\mid−{ln}\mid{b}\mid\right) \\ $$$${ln}\mid{a}\mid={ln}\mid{x}\mid−{ln}\mid{a}\mid−{ln}\mid{x}\mid+{ln}\mid{b}\mid \\ $$$$\mathrm{2}\centerdot{ln}\mid{a}\mid={ln}\mid{b}\mid \\ $$$${e}^{{ln}\mid{a}^{\mathrm{2}} \mid} ={e}^{{ln}\mid{b}\mid} \\ $$$${a}^{\mathrm{2}} ={b} \\ $$$$\mathrm{if}\:{a}^{\mathrm{2}} ={b}\:\mathrm{then} \\ $$$${a}=\frac{{x}−{a}}{{x}−{b}} \\ $$$${a}\centerdot\left({x}−{b}\right)=\left({x}−{a}\right)\Leftrightarrow{x}−{b}\neq\mathrm{0} \\ $$$${a}\centerdot{x}−{a}\centerdot{b}={x}−{a} \\ $$$${a}\centerdot{x}−{x}={a}\centerdot{b}−{a} \\ $$$${x}\left({a}−\mathrm{1}\right)={a}\centerdot{a}^{\mathrm{2}} −{a} \\ $$$${x}=\frac{{a}^{\mathrm{3}} −{a}}{{a}−\mathrm{1}}\Leftrightarrow{a}−\mathrm{1}\neq\mathrm{0} \\ $$$${x}=\frac{{a}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}{{a}−\mathrm{1}} \\ $$$${x}=\frac{{a}\centerdot\left({a}−\mathrm{1}\right)\centerdot\left({a}+\mathrm{1}\right)}{{a}−\mathrm{1}} \\ $$$${x}={a}\left({a}+\mathrm{1}\right) \\ $$$${x}={b}+{a} \\ $$$$ \\ $$$$\mathrm{answer}: \\ $$$${x}={a}^{\mathrm{2}} +{a}\:{or}\:{x}={b}+{a},\:\mathrm{and}\:{x},{a},{b}\:\notin\:\mathbb{C} \\ $$$$ \\ $$$$\mathrm{so}\:\mathrm{now}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}:\:\mathrm{what}\:\mathrm{if}\:{x},{a},{b}\:\in\:\mathbb{C}? \\ $$$$ \\ $$

Question Number 191250 Answers: 0 Comments: 4

Question Number 191243 Answers: 2 Comments: 1

Question Number 191232 Answers: 1 Comments: 0

If x^2 − y^2 = 2023^(2023) then how many pair of x,y where x, y ∈ N

$$\mathrm{If}\:{x}^{\mathrm{2}} \:−\:{y}^{\mathrm{2}} \:=\:\mathrm{2023}^{\mathrm{2023}} \:\mathrm{then}\:\mathrm{how}\:\mathrm{many}\: \\ $$$$\mathrm{pair}\:\mathrm{of}\:{x},{y}\:{where}\:{x},\:{y}\:\in\:\mathrm{N} \\ $$

Question Number 191231 Answers: 0 Comments: 1

Logic puzzle if 3621+22=10 7842+10=17 8223+13=29 1930+7 = 10 6624+11=?

$${Logic}\:{puzzle} \\ $$$${if}\:\:\mathrm{3621}+\mathrm{22}=\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\mathrm{7842}+\mathrm{10}=\mathrm{17} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{8223}+\mathrm{13}=\mathrm{29} \\ $$$$\:\:\:\:\:\:\:\mathrm{1930}+\mathrm{7}\:=\:\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\mathrm{6624}+\mathrm{11}=? \\ $$

Question Number 191283 Answers: 1 Comments: 0

Question Number 191222 Answers: 0 Comments: 0

Question Number 191221 Answers: 0 Comments: 0

Question Number 191220 Answers: 1 Comments: 5

x^6 −x^3 =2 solve for x m=x^3 m^2 =(x^3 )^2 m^2 −m=2 m^2 −m−2=0 α+β=1→α=(1/2)+μ and β=(1/2)−μ α∙β=((1/2)+μ)∙((1/2)−μ)=−2 α∙β=((1/2)−μ)∙((1/2)+μ)=−2 (1/4)−μ^2 =−2 μ^2 =2+(1/4)=(9/4) μ=(√(9/4))=(3/2) α=(1/2)+(3/2)=((−1+3)/2) β=(1/2)−(3/2)=((−1−3)/2) (m+((1+3)/2))(m+((1−3)/2))=0 m_1 =((1+3)/2) m_2 =−((1−3)/2) m_(1,2) =((1±3)/2) m=x^3 →x=(((1±3)/2))^(1/3)

$${x}^{\mathrm{6}} −{x}^{\mathrm{3}} =\mathrm{2}\:{solve}\:{for}\:{x} \\ $$$$ \\ $$$${m}={x}^{\mathrm{3}} \\ $$$${m}^{\mathrm{2}} =\left({x}^{\mathrm{3}} \right)^{\mathrm{2}} \\ $$$${m}^{\mathrm{2}} −{m}=\mathrm{2} \\ $$$${m}^{\mathrm{2}} −{m}−\mathrm{2}=\mathrm{0} \\ $$$$\alpha+\beta=\mathrm{1}\rightarrow\alpha=\frac{\mathrm{1}}{\mathrm{2}}+\mu\:\:{and}\:\:\beta=\frac{\mathrm{1}}{\mathrm{2}}−\mu \\ $$$$\alpha\centerdot\beta=\left(\frac{\mathrm{1}}{\mathrm{2}}+\mu\right)\centerdot\left(\frac{\mathrm{1}}{\mathrm{2}}−\mu\right)=−\mathrm{2} \\ $$$$\alpha\centerdot\beta=\left(\frac{\mathrm{1}}{\mathrm{2}}−\mu\right)\centerdot\left(\frac{\mathrm{1}}{\mathrm{2}}+\mu\right)=−\mathrm{2} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}−\mu^{\mathrm{2}} =−\mathrm{2} \\ $$$$\mu^{\mathrm{2}} =\mathrm{2}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\mu=\sqrt{\frac{\mathrm{9}}{\mathrm{4}}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\alpha=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}=\frac{−\mathrm{1}+\mathrm{3}}{\mathrm{2}} \\ $$$$\beta=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}=\frac{−\mathrm{1}−\mathrm{3}}{\mathrm{2}} \\ $$$$\left({m}+\frac{\mathrm{1}+\mathrm{3}}{\mathrm{2}}\right)\left({m}+\frac{\mathrm{1}−\mathrm{3}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$${m}_{\mathrm{1}} =\frac{\mathrm{1}+\mathrm{3}}{\mathrm{2}} \\ $$$${m}_{\mathrm{2}} =−\frac{\mathrm{1}−\mathrm{3}}{\mathrm{2}} \\ $$$${m}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{1}\pm\mathrm{3}}{\mathrm{2}} \\ $$$${m}={x}^{\mathrm{3}} \rightarrow{x}=\left(\frac{\mathrm{1}\pm\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$ \\ $$

Question Number 191430 Answers: 3 Comments: 0

factorise 4x^4 +81

$${factorise}\:\mathrm{4}{x}^{\mathrm{4}} +\mathrm{81} \\ $$

Question Number 191429 Answers: 0 Comments: 0

Best prime generating formula

$${Best}\:{prime}\:{generating}\:{formula} \\ $$

Question Number 191229 Answers: 0 Comments: 0

evaluate P_n =(a+sin θ)(a+sin (θ/2))(a+sin (θ/2^2 ))...(a+sin (θ/2^n )) with ∣a∣≤1

$${evaluate} \\ $$$${P}_{{n}} =\left({a}+\mathrm{sin}\:\theta\right)\left({a}+\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\right)\left({a}+\mathrm{sin}\:\frac{\theta}{\mathrm{2}^{\mathrm{2}} }\right)...\left({a}+\mathrm{sin}\:\frac{\theta}{\mathrm{2}^{{n}} }\right) \\ $$$${with}\:\mid{a}\mid\leqslant\mathrm{1} \\ $$

Question Number 191212 Answers: 0 Comments: 0