Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
Just take both sides to the power of $-1/0.45$. –  Sanath Devalapurkar Feb 10 at 2:45
add comment

4 Answers

up vote 3 down vote accepted

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 $$x=\sqrt[9]{\left(\frac{8}{y}\right)^{20}}$$

share|improve this answer
    
thank you!!!!!!!! –  Jessica Feb 10 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 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 at 2:07
add comment

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

share|improve this answer
add comment

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

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.