1. ## Evaluate

evaluate $(\sqrt[3]{0.216})(243^{-\frac{2}{5}})$

I am not too sure if I am on the right track...

Attempt:

$(\sqrt[3]{0.216})(243^{-\frac{2}{5}})$

= $\frac{\sqrt[3]{0.216}}{243^{\frac{2}{5}}}$

= $\sqrt[3]{\frac{0.216}{243^{\frac{6}{5}}}}$

2. ## evaluate

Originally Posted by Punch
evaluate $(\sqrt[3]{0.216})(243^{-\frac{2}{5}})$

I am not too sure if I am on the right track...

Attempt:

$(\sqrt[3]{0.216})(243^{-\frac{2}{5}})$

= $\frac{\sqrt[3]{0.216}}{243^{\frac{2}{5}}}$

= $\sqrt[3]{\frac{0.216}{243^{\frac{6}{5}}}}$

first step is fine. Answer using calculator or logs. The numerator is .6 and the denominator is 9.

bjh

3. Originally Posted by Punch
evaluate $(\sqrt[3]{0.216})(243^{-\frac{2}{5}})$

I am not too sure if I am on the right track...

Attempt:

$(\sqrt[3]{0.216})(243^{-\frac{2}{5}})$

= $\frac{\sqrt[3]{0.216}}{243^{\frac{2}{5}}}$

= $\sqrt[3]{\frac{0.216}{243^{\frac{6}{5}}}}$
Hi Punch,

$243=3^5$

$243^{\frac{1}{5}}=3$

$243^{\frac{6}{5}}=3^6$

$0.216=\frac{216}{1000}=\frac{6^3}{10^3}=\frac{2^33 ^3}{2^35^3}=\frac{3^3}{5^3}$

$\sqrt[3]{\frac{0.216}{243^{\frac{6}{5}}}}=\sqrt[3]{\frac{3^3}{5^33^6}}=\sqrt[3]{\frac{1}{5^33^3}}=\frac{1}{5(3)}=\frac{1}{15}$

4. is it necessary to simplify it so much?

5. ## evaluate

Originally Posted by Punch
is it necessary to simplify it so much?

Yes if you were asked to solve without logs or calculator

bjh