# Thread: solving for x, involving logs

1. ## 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)

2. 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?

$\displaystyle y-9=\sqrt[4]{x}\color{red}{{-}}$$\displaystyle \sqrt[5]{x-1.5}$?

3. No

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

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

4. 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

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

5. 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.

6. Can someone tell me where to go from there, or if that's right?

7. 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: $\displaystyle x \approx 5.2926$