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)
....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?
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: $\displaystyle x \approx 5.2926$