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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- Jun 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) - Jun 7th 2008, 06:37 PMMathstud28
- Jun 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 - Jun 7th 2008, 06:51 PMMathstud28
....uhm....I think you mean that you have two simultaneous equations, namely

$\displaystyle y=\sqrt[5]{x}+9$

and $\displaystyle y=\sqrt[4]{x-1.5}+9$

Now seeing that if we subtract nine from both we can set them equal to each other getting

$\displaystyle \sqrt[5]{x}=\sqrt[4]{x-1.5}$

Now how are you proposing we solve this? - Jun 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. - Jun 8th 2008, 11:31 AMtheowne
Can someone tell me where to go from there, or if that's right?

- Jun 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:

$\displaystyle x^{0.2} = (x-1.5)^{0.25}$

$\displaystyle x^{\frac{1}{5}} = (x - 1.5)^{\frac{1}{4}}$

$\displaystyle x^{4} = (x - 1.5)^{5}$ (Raised both sides to the power of 20)

$\displaystyle x^{4} = x^{5} + 5x^{4}(-1.5) + 10x^{3}(-1.5)^{2} + 10x^{2}(-1.5)^{3} + 5x(-1.5)^{4} + (-1.5)^{5}$ (via binomial theorem)

$\displaystyle 0 = x^{5} - 8.5x^{4} + 22.5x^{3} - 33.75x^{2} + 25.3125x - 7.59375$

You would somehow have to find the roots of that quintic equation, keeping in mind that $\displaystyle x-1.5\geq 0$

There seems to be one real root: $\displaystyle x \approx 5.2926$