Question

How do I rearrange this equation in terms of x?

Answer 1

Let me type it out, hold on!

Answer 2

\[k*\ln \frac{ cL }{ c}=\frac{ Q }{ A}(1-x)-k(x-1)\]

Answer 3

I should get this

Answer 4

@hartnn any ideas?!

Answer 5

My algebra skills letting me down again!

Answer 6

@dan815 any ideas!?

Answer 7

k∗lncLc=QA(1−x)−k(x−1)

k*ln cL /c = Q/A (1-x) -k(x-1)

k*ln cL/c = Q/A - xQ/A - kx +k

k*ln cL/c -Q/A -k = x(-k-Q/A)
kln cL/c - Q/A -k
x = ----------------------
-k- Q/A

Ak*ln cL/c - Q - Ak
--------------------
A Ak ln cL/c -Q -Ak -Ak -Q
x = ------------------------- = ------------------ = -------- - ----------
- Ak -Q -Ak -Q -Ak -Q
---------
A


-Ak -Q Ak ln cL/c Ak ln cL/c ln cL/c
= -------- - ---------- = 1 - ---------- 1 - -----------
-Ak -Q Ak +Q Ak +Q 1 + Q/Ak

so yes i think you are right

hope this will help you

good luck

Answer 8

sorry i missed one equality sign



-Ak -Q Ak ln cL/c Ak ln cL/c ln cL/c
= -------- - ---------- = 1 - ---------- = 1 - -----------
-Ak -Q Ak +Q Ak +Q 1 + Q/Ak

hope now is clearly