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Tombrokaw
What is the simplest way to do this? 49^-3/2
Well, you can simplify \[49^{-\frac{ 3 }{ 2 }}\] to \[\left( 49^{\frac{ 1 }{ 2 }} \right)^{-3}\] And\[49^{\frac{ 1 }{ 2 }}=\sqrt{49}=7\] so \[\left( 49^{\frac{ 1 }{ 2 }} \right)^{-3}=\left( 7 \right)^{-3}=\frac{ 1 }{ 7\times7\times7 }\] This means that\[ 49^{-\frac{ 3 }{ 2 }}=\frac{ 1 }{ 7\times7\times7 }\] If anything about this seems to complicated, let me know. The bottom line is that you're splitting the exponent into a power of 1/2 and a power of -3.
Yes I was able to get to the answer with help of another system actually but I thought it was too complicated and I was wondering if there was a much simpler way to do this?
In 49^1/2 . the denominator tells you if its a cube root or a square root or a quad root ?
For fractional exponents, it's power-over-root.
Oh alright. Thank you. That clarifies it.