Just wondering if this is the case: $\displaystyle \displaystyle 3^\frac{m}{n}=\frac{m}{3^n}$ The exponent in the first term should be negative but i don't know how to do that in latex sorry! Thanks BIOS
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$\displaystyle 3^{-\frac{m}{n}} = \frac{1}{3^{\frac{m}{n}}}$ click on the formula to see the latex.
Originally Posted by BIOS Just wondering if this is the case: $\displaystyle \displaystyle 3^\frac{m}{n}=\frac{m}{3^n}$ The exponent in the first term should be negative but i don't know how to do that in latex sorry! Thanks BIOS $\displaystyle \displaystyle\frac{m}{3^n}=m\left(\frac{1}{3^n}\ri ght)=m\left(\frac{3^0}{3^n}\right)=m\left(3^{0-n}\right)=m3^{-n}$
Thanks alot for the replies and the clarification. So is this the case: $\displaystyle 3^{-\frac{m}{n}} = \frac{1}{3^{\frac{m}{n}}}=\frac{1}{\sqrt [n]{3^m}}$
Originally Posted by BIOS Thanks alot for the replies and the clarification. So is this the case: $\displaystyle 3^{-\frac{m}{n}} = \frac{1}{3^{\frac{m}{n}}}=\frac{1}{\sqrt [n]{3^m}}$ Yup.
Sweet. Thanks again for the help. I have most of the rules down now. Just need to work on my logarithms!
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