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Question Number 191254 by anr0h3 last updated on 21/Apr/23
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?
$$ \\ $$$${a}=\frac{{x}−{a}}{{x}−{b}}\:{solve}\:{for}\:{x} \\ $$$$\mathrm{did}\:\mathrm{i}\:\mathrm{make}\:\mathrm{anything}\:\mathrm{wrong}\:\mathrm{in}\:\mathrm{the}\:\mathrm{following}? \\ $$$$ \\ $$$$\mathrm{starpoint}: \\ $$$${ln}\mid{a}\mid={ln}\mid\frac{{x}−{a}}{{x}−{b}}\mid \\ $$$${ln}\mid{a}\mid={ln}\mid{x}−{a}\mid−{ln}\mid{x}−{b}\mid \\ $$$${ln}\mid{a}\mid={ln}\mid\frac{{x}}{{a}}\mid−{ln}\mid\frac{{x}}{{b}}\mid \\ $$$${ln}\mid{a}\mid={ln}\mid{x}\mid−{ln}\mid{a}\mid−\left({ln}\mid{x}\mid−{ln}\mid{b}\mid\right) \\ $$$${ln}\mid{a}\mid={ln}\mid{x}\mid−{ln}\mid{a}\mid−{ln}\mid{x}\mid+{ln}\mid{b}\mid \\ $$$$\mathrm{2}\centerdot{ln}\mid{a}\mid={ln}\mid{b}\mid \\ $$$${e}^{{ln}\mid{a}^{\mathrm{2}} \mid} ={e}^{{ln}\mid{b}\mid} \\ $$$${a}^{\mathrm{2}} ={b} \\ $$$$\mathrm{if}\:{a}^{\mathrm{2}} ={b}\:\mathrm{then} \\ $$$${a}=\frac{{x}−{a}}{{x}−{b}} \\ $$$${a}\centerdot\left({x}−{b}\right)=\left({x}−{a}\right)\Leftrightarrow{x}−{b}\neq\mathrm{0} \\ $$$${a}\centerdot{x}−{a}\centerdot{b}={x}−{a} \\ $$$${a}\centerdot{x}−{x}={a}\centerdot{b}−{a} \\ $$$${x}\left({a}−\mathrm{1}\right)={a}\centerdot{a}^{\mathrm{2}} −{a} \\ $$$${x}=\frac{{a}^{\mathrm{3}} −{a}}{{a}−\mathrm{1}}\Leftrightarrow{a}−\mathrm{1}\neq\mathrm{0} \\ $$$${x}=\frac{{a}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}{{a}−\mathrm{1}} \\ $$$${x}=\frac{{a}\centerdot\left({a}−\mathrm{1}\right)\centerdot\left({a}+\mathrm{1}\right)}{{a}−\mathrm{1}} \\ $$$${x}={a}\left({a}+\mathrm{1}\right) \\ $$$${x}={b}+{a} \\ $$$$ \\ $$$$\mathrm{answer}: \\ $$$${x}={a}^{\mathrm{2}} +{a}\:{or}\:{x}={b}+{a},\:\mathrm{and}\:{x},{a},{b}\:\notin\:\mathbb{C} \\ $$$$ \\ $$$$\mathrm{so}\:\mathrm{now}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}:\:\mathrm{what}\:\mathrm{if}\:{x},{a},{b}\:\in\:\mathbb{C}? \\ $$$$ \\ $$
Commented by mr W last updated on 23/Apr/23
first: ln (a−b)≠ln (a/b), therefore ln (a−b)≠ln a−ln b second: if a=(b/c), then ac=b third: when x=a^2 +a or x=b+a, why should x,a,b ∉C? fourth: when somebody even doesn′t know a=(b/c) ⇒ ac=b, how can he understand complex numbers? etc.
$${first}: \\ $$$$\mathrm{ln}\:\left({a}−{b}\right)\neq\mathrm{ln}\:\frac{{a}}{{b}},\:{therefore} \\ $$$$\mathrm{ln}\:\left({a}−{b}\right)\neq\mathrm{ln}\:{a}−\mathrm{ln}\:{b} \\ $$$$ \\ $$$${second}: \\ $$$${if}\:{a}=\frac{{b}}{{c}},\:{then}\:{ac}={b} \\ $$$$ \\ $$$${third}: \\ $$$${when}\:{x}={a}^{\mathrm{2}} +{a}\:{or}\:{x}={b}+{a},\:{why} \\ $$$${should}\:{x},{a},{b}\:\notin{C}? \\ $$$$ \\ $$$${fourth}: \\ $$$${when}\:{somebody}\:{even}\:{doesn}'{t}\:{know}\: \\ $$$${a}=\frac{{b}}{{c}}\:\Rightarrow\:{ac}={b},\:{how}\:{can}\:{he}\:{understand} \\ $$$${complex}\:{numbers}? \\ $$$$ \\ $$$${etc}. \\ $$
Commented by mr W last updated on 21/Apr/23
therefore a=((x−a)/(x−b)) a(x−b)=x−a ax−ab=x−a (a−1)x=a(b−1) x=((a(b−1))/(a−1)) (a≠1)
$${therefore} \\ $$$${a}=\frac{{x}−{a}}{{x}−{b}} \\ $$$${a}\left({x}−{b}\right)={x}−{a} \\ $$$${ax}−{ab}={x}−{a} \\ $$$$\left({a}−\mathrm{1}\right){x}={a}\left({b}−\mathrm{1}\right) \\ $$$${x}=\frac{{a}\left({b}−\mathrm{1}\right)}{{a}−\mathrm{1}}\:\:\:\left({a}\neq\mathrm{1}\right) \\ $$
Commented by JDamian last updated on 21/Apr/23
you seem to misunderstand the exponential function properties
Commented by mehdee42 last updated on 21/Apr/23
ln∣((x−a)/a)∣−ln∣x−b∣=0⇏e^(ln∣((x−a)/a)∣) −e^(ln∣x−b∣) =1
$${ln}\mid\frac{{x}−{a}}{{a}}\mid−{ln}\mid{x}−{b}\mid=\mathrm{0}\nRightarrow{e}^{{ln}\mid\frac{{x}−{a}}{{a}}\mid} −{e}^{{ln}\mid{x}−{b}\mid} =\mathrm{1} \\ $$
Commented by JDamian last updated on 21/Apr/23
e^(x−y) ≠ e^x −e^y
$$\mathrm{e}^{\mathrm{x}−\mathrm{y}} \:\neq\:\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{\mathrm{y}} \\ $$
Commented by anr0h3 last updated on 21/Apr/23
ok thanks

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