## ursularoxursox 2 years ago Please help me out. I should be able to do the rest on my own once I see the steps to solve this problem. (64x^6)^(-5/3) I have to simplify

1. ZeHanz

You need to know the basic rules of powers:$a^{-b}=\frac{ 1 }{ a^b }$$a^{\frac{ ^b }{ c }} =\sqrt[c]{a^b}=(\sqrt[c]{a})^b$$(a^b)^c=a^{bc}$$a^b \cdot a^c=a^{b+c}$$\frac{ a^b }{ a^c }=a^{b-c}$ Try to see what you need to do with these rules here...

2. ursularoxursox

alright i did, thanks, this is what I got. $\sqrt[5]{64x ^{26}}$

3. ursularoxursox

5 root 64x^26 in case the other post was hard to read

4. ZeHanz

I got to: $(64x^6)^{-\frac{ 5 }{ 3 }}=\frac{ 1 }{ (64x^6)^{\frac{ 5 }{ 3 } }}=\frac{ 1 }{ 64^{\frac{5}{3}}x^{6 \cdot \frac{5}{3}} }=\frac{ 1 }{64^{\frac{5}{3}} x^{10}}$Now we have that nasy 64^(5/3)... But 64=2^6! so we can go a bit further:$\frac{ 1 }{ (2^6)^{\frac{5}{3}} x^{10}}=\frac{ 1 }{ 2^{10}x^{10} }=\frac{ 1 }{ 1024x^{10} }$

5. ursularoxursox

thank you so much! it makes sense. I just have to remember my exponent rules and factors :) thanks for your help!

6. ZeHanz

Once you've seen it again, you'll remember...just practise a few more and you're safe!