# Solve $y = 8x^{-0.45}$ in terms of $x$?

Attempt:

$\displaystyle\frac{y}{8} = x^{-0.45}$

How do I do the negative root so I have this equation in terms of $x$, since this is close to $.5$ so it should be the square root?

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Just take both sides to the power of $-1/0.45$. – Sanath K. Devalapurkar Feb 10 '14 at 2:45

If you are averse to negative exponents, they just indicate a reciprocal. You have $$\frac{y}{8}=\frac{1}{x^{0.45}}$$ So then $$x^{0.45}=\frac{8}{y}$$ or $$x^{9/20}=\frac{8}{y}$$ and then $$x^{9}=\left(\frac{8}{y}\right)^{20}$$ giving $$x=\sqrt[9]{\left(\frac{8}{y}\right)^{20}}$$

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thank you!!!!!!!! – Jessica Feb 10 '14 at 3:05
@Jessica: Don't be put off by what looks like a complicated exponent. Remember, it's just a number and so in theory it isn't any harder than solving $y=x^5$. Generally, look for the pattern behind the form. So if you have an equation of the form $y=Ax^B$ (where $A$ and $B$ are numbers), divide by $A$ to get $y/A=x^B$. Then raise both sides to the power $1/B$ to get your answer $(y/A)^{1/B}= x$. – MPW Feb 10 '14 at 3:46
@MPW Thank you so much for the insight! Yes i must admit, the big kid numbers tend to scare me away haha but thank you!!! – Jessica Feb 11 '14 at 2:07

No its only CLOSE to the square root. But remember $x^{-n}$ = $1/x^{n}$ then $y/8=x^{-.45}.... y=8/x^{.45}$

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$\frac{y}{8}=x^{-0.45}\Rightarrow \frac{1}{x}=(\frac{y}{8})^{1/.45}\Rightarrow \frac{1}{x}=(\frac{y}{8})^{20/9}\Rightarrow x=(\frac{8}{y})^{20/9}$ provided $x,y\not=0$.

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Use log,

$y/8=x^{-0,45}$

$log(y/8)=-0,45log(x)$

$log((y/8)^{1/-0,45})=log(x)$

$(y/8)^{1/-0,45}=x$

Use $log(x^y)=ylog(x)$

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