For values of x which a solution exists solve for y

[steve snash]

Homework Statement

(For those values of x for which a solution exists), solve the following equation for y
3e^(2y−8) = (2x^5)−3

Homework Equations

1. xln e= x

The Attempt at a Solution

1.3e^(2y−8) = (2x^5)−3
2.e^(2y−8) = ((2x^5)−3)/3
3.(2y-8)ln e = (ln (2x^5)-3)/3
4.(2y-8) = (ln (2x^5)-3)/3
5.y= ((ln (2x^5)-3/6)+8

this answer was wrong just wondering where i went wrong??

 

[Dustinsfl]

$$e^{2y-8}=\frac{2x^{5}}{3}-1$$

$$2y-8=ln({\frac{2x^{5}}{3}-1})$$

$$2y=ln(\frac{2x^{5}}{3}-1)+8$$

$$y=\frac{ln(\frac{2x^{5}}{3}-1)}{2}+4$$

 

[steve snash]

Thanks man