Loading [MathJax]/jax/output/SVG/jax.js
web2.0calc
 
+0  
 

 y = (e^6x + e^−6x)/ (e^6x − e^−6x)

0
1115
3
avatar

y = (e^6x + e^−6x)/(e^6x − e^−6x)

 

Solve for x in terms of y

 Jul 6, 2015

Best Answer 

 #1
avatar
+10

y=(e(6×x)+1(e(6×x)))(e(6×x)−1(e(6×x)))

y=((e(12×x)+1)e(6×x))((e(12×x)−1)e(6×x))

y=(e(12×x)+1)(e(12×x)−1)

y=(e(12×x)−1)(e(12×x)−1)+2(e(12×x)−1)

y=1+2(e(12×x)−1)

y−1=2(e(12×x)−1)

((y−1)×(e(12×x)−1))=2

e(12×x)−1=2(y−1)

e(12×x)=2(y−1)+1

e(12×x)=(y+1)(y−1)

12×x=ln(y+1)×ln(y−1)

x=(ln(y+1)×ln(y−1))12

.
 Jul 6, 2015

3+2 Answers

 #1
avatar
+10
Best Answer

y=(e(6×x)+1(e(6×x)))(e(6×x)−1(e(6×x)))

y=((e(12×x)+1)e(6×x))((e(12×x)−1)e(6×x))

y=(e(12×x)+1)(e(12×x)−1)

y=(e(12×x)−1)(e(12×x)−1)+2(e(12×x)−1)

y=1+2(e(12×x)−1)

y−1=2(e(12×x)−1)

((y−1)×(e(12×x)−1))=2

e(12×x)−1=2(y−1)

e(12×x)=2(y−1)+1

e(12×x)=(y+1)(y−1)

12×x=ln(y+1)×ln(y−1)

x=(ln(y+1)×ln(y−1))12

Guest Jul 6, 2015
 #2
avatar+23254 
+5

In the second-to-the-last line, shouldn't the times sign be a minus sign?

That is:  12x  =  ln(y + 1)  -  ln(y - 1)

                x  =  [ ln(y + 1)  -  ln(y - 1) ] / 12

 Jul 6, 2015
 #3
avatar
+5

Yes, it should be. I made an arithmetic error.

 Jul 7, 2015

0 Online Users

Top Users

avatar+130458 CPhill moderator
avatar+37165 ElectricPavlov
avatar+33654 Alan moderator
avatar+15052 asinus moderator
avatar+3146 admin administrator
avatar+2249 GingerAle
avatar+1776 bader
avatar+1734 blackpanther
avatar+1721 parmen
avatar+1556 kittykat
avatar+1493 learnmgcat

Sticky Topics