Long for help with (x+4)^3/5=(5x-1)^2/3

skyscraper_ming said:
(x+4)^3/5=(5x-1)^2/3

I have been spending hours on it.Please help me.Thanks! :D
Click to expand...
Then you should have plenty to show! :wink:

Please reply with a clear listing of your work and reasoning so far. When you reply, please clarify the exponents / divisors; in particular, do you mean [(x + 4)^3]/5 or (x + 4)^(3/5) or something else, for the left-hand side of your equation? (Same question on the right-hand side of the equation.)

Please be complete. Thank you! :D

Eliz.
 
What I mean is that I would like someone to help me solve this problem because I could not figure it out even I spent a lot of time on it.Anyone please?
 
You received exactly what you requested. It's hard to judge what is needed without a single hint concerning what you have tried.

Have you tried raising both sides to the 15th power?

Anything?! Throw us a bone.
 
I think you are on a wild goose chase.
I guess you mean \(\displaystyle (x+4)^{\frac{3}{5}}=(5x-1)^{\frac{2}{3}}\)
although literal interpretation of the symbols means \(\displaystyle \frac{(x+4)^3}{5}=\frac{(5x-1)^2}{3}\)
Try solving this last one.
For the former...
Let x+4=u and 5x-1=v and raise both sides to the 15th.
\(\displaystyle (u^{3/5})^{15} = (v^{2/3})^{15}\)
\(\displaystyle u^9 = v^{10}\)
The only possibilities for this to be true is u=v=0 or u=v=1.
Neither of those will work.
I'm probably missing something.
In what course of math did this problem arise?
 
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