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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