solving for x, involving logs

• Jun 7th 2008, 07:35 PM
theowne
solving 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, 07:37 PM
Mathstud28
Quote:

Originally Posted by theowne
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)

What is this?

$y-9=\sqrt[4]{x}\color{red}{{-}}$ $\sqrt[5]{x-1.5}$?
• Jun 7th 2008, 07:43 PM
theowne
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, 07:51 PM
Mathstud28
Quote:

Originally Posted by theowne
No

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

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

....uhm....I think you mean that you have two simultaneous equations, namely

$y=\sqrt[5]{x}+9$
and $y=\sqrt[4]{x-1.5}+9$

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

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

Now how are you proposing we solve this?
• Jun 7th 2008, 07:54 PM
theowne
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, 12:31 PM
theowne
Can someone tell me where to go from there, or if that's right?
• Jun 8th 2008, 01:13 PM
o_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:
$x^{0.2} = (x-1.5)^{0.25}$
$x^{\frac{1}{5}} = (x - 1.5)^{\frac{1}{4}}$
$x^{4} = (x - 1.5)^{5}$ (Raised both sides to the power of 20)
$x^{4} = x^{5} + 5x^{4}(-1.5) + 10x^{3}(-1.5)^{2} + 10x^{2}(-1.5)^{3} + 5x(-1.5)^{4} + (-1.5)^{5}$ (via binomial theorem)
$0 = x^{5} - 8.5x^{4} + 22.5x^{3} - 33.75x^{2} + 25.3125x - 7.59375$

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

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