Printable View
Hello,
To evaluate the expression (x-2)^1/2 the next steps listed are x^(-2)(1/2)
then x^-1 and finally 1/x. I am fine with the final two steps.
Please help me see how or what rules are used to go from the first to second step. Is this just a basic rules I am forgetting?
So in summary how does (x-2)^1/2 become x^(-2)(1/2)
Thank you.
1. An example:Quote:
Originally Posted by joseclar
$\displaystyle \left(b^3\right)^4 = \left(b^3\right) \cdot \left(b^3\right) \cdot \left(b^3\right) \cdot \left(b^3\right) = \left(b\right)^{3+3+3+3} = b^{3 \cdot 4}$
So if you take a power to a power you have to multiply the exponents.
2. In general:
$\displaystyle \left(b^a\right)^c = b^{a\cdot c} = \left(b^c\right)^a$
3. Apply this rule to your question.
tHANK YOU for the reply. I am aware of the rule for multiplying exponents. My question is how does (x-2)^1/2 become x^-2(1/2). Thanks.
Why do you think it does? In what context does it appear?Quote:
Originally Posted by joseclar
... by typo!Quote:
Originally Posted by joseclar
1. I assumed that there was a typo in the original term and that actually was meant $\displaystyle \left(x^{-2} \right)^{\frac12}$Quote:
Originally Posted by joseclar
2. If you re-arrange the given term
$\displaystyle (x-2)^{\frac12} = \sqrt{x-2}$
you can see that
$\displaystyle \sqrt{x-2} = \frac1x~\implies~x^3-2x^2-1=0$
has only one real solution at $\displaystyle x = \frac{\sqrt[3]{172-12 \cdot \sqrt{177}}}{6}+\frac{\sqrt[3]{172+12 \cdot \sqrt{177}}}{6} +\frac23$
To answer your question So in summary how does (x-2)^1/2 become x^(-2)(1/2): Only once - and in all other cases this transformation is wrong.
