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?


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 191254 by anr0h3 last updated on 21/Apr/23

$a = \frac{x - a}{x - b} s o l v e f o r x$ $did i make anything wrong in the following ?$ $starpoint :$ $l n ∣ a ∣= l n ∣ \frac{x - a}{x - b} ∣$ $l n ∣ a ∣= l n ∣ x - a ∣ - l n ∣ x - b ∣$ $l n ∣ a ∣= l n ∣ \frac{x}{a} ∣ - l n ∣ \frac{x}{b} ∣$ $l n ∣ a ∣= l n ∣ x ∣ - l n ∣ a ∣ - (l n ∣ x ∣ - l n ∣ b ∣)$ $l n ∣ a ∣= l n ∣ x ∣ - l n ∣ a ∣ - l n ∣ x ∣ + l n ∣ b ∣$ $2 \cdot l n ∣ a ∣= l n ∣ b ∣$ $e^{l n ∣ a^{2} ∣} = e^{l n ∣ b ∣}$ $a^{2} = b$ $if a^{2} = b then$ $a = \frac{x - a}{x - b}$ $a \cdot (x - b) = (x - a) \Leftrightarrow x - b \neq 0$ $a \cdot x - a \cdot b = x - a$ $a \cdot x - x = a \cdot b - a$ $x (a - 1) = a \cdot a^{2} - a$ $x = \frac{a^{3} - a}{a - 1} \Leftrightarrow a - 1 \neq 0$ $x = \frac{a (a^{2} - 1)}{a - 1}$ $x = \frac{a \cdot (a - 1) \cdot (a + 1)}{a - 1}$ $x = a (a + 1)$ $x = b + a$ $answer :$ $x = a^{2} + a o r x = b + a, and x, a, b \notin C$ $so now the question is : what if x, a, b \in C ?$
Commented by mr W last updated on 23/Apr/23

$f i r s t :$ $\ln (a - b) \neq \ln \frac{a}{b}, t h e r e f o r e$ $\ln (a - b) \neq \ln a - \ln b$ $s e c o n d :$ $i f a = \frac{b}{c}, t h e n a c = b$ $t h i r d :$ $w h e n x = a^{2} + a o r x = b + a, w h y$ $s h o u l d x, a, b \notin C ?$ $f o u r t h :$ $w h e n s o m e b o d y e v e n {d o e s n}^{'} t k n o w$ $a = \frac{b}{c} \Rightarrow a c = b, h o w c a n h e u n d e r s t a n d$ $c o m p l e x n u m b e r s ?$ $e t c .$
Commented by mr W last updated on 21/Apr/23

$t h e r e f o r e$ $a = \frac{x - a}{x - b}$ $a (x - b) = x - a$ $a x - a b = x - a$ $(a - 1) x = a (b - 1)$ $x = \frac{a (b - 1)}{a - 1} (a \neq 1)$
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

$l n ∣ \frac{x - a}{a} ∣ - l n ∣ x - b ∣= 0 ⇏ e^{l n ∣ \frac{x - a}{a} ∣} - e^{l n ∣ x - b ∣} = 1$
Commented by JDamian last updated on 21/Apr/23

$e^{x - y} \neq e^{x} - e^{y}$
Commented by anr0h3 last updated on 21/Apr/23
ok thanks
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum