y-9 = x^0.25 = (x-1.5)^0.2

How can I solve this? I guess logs would come in at some point. I tried rearranging a bit but then got this

0.25/0.2 = logx / log(x-1.5)

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- June 7th 2008, 06:35 PMtheownesolving for x, involving logs
y-9 = x^0.25 = (x-1.5)^0.2

How can I solve this? I guess logs would come in at some point. I tried rearranging a bit but then got this

0.25/0.2 = logx / log(x-1.5) - June 7th 2008, 06:37 PMMathstud28
- June 7th 2008, 06:43 PMtheowne
No

y=(x)^0.2+9

y=(x-1.5)^0.25+9

y-9=x^0.2=(x-1.5)^0.25 - June 7th 2008, 06:51 PMMathstud28
- June 7th 2008, 06:54 PMtheowne
I do not know. I tried to do log of both sides and get

0.25 log (x-1.5) = 0.2 log x

0.25/0.2 = log x / log (x-1.5)

Don't know what to do further or if that's even right. - June 8th 2008, 11:31 AMtheowne
Can someone tell me where to go from there, or if that's right?

- June 8th 2008, 12:13 PMo_O
If this is indeed your question, I don't think logarithms will help you much. There's no way in isolating

**x**from log(x - 1.5) so you can't do all that much after what you've done.

If you were to attempt to solve it:

(Raised both sides to the power of 20)

(via binomial theorem)

You would somehow have to find the roots of that quintic equation, keeping in mind that

There seems to be one real root: