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Evaluating (indices) help

I dont understand how to evaluate this:

( 9^10 / 9^4 )^1/3

Can someone explain how to get the answer of 81?
Reply 1
Avatar for Notnek
Notnek
21
Original post by ozmo19
I dont understand how to evaluate this:

( 9^10 / 9^4 )^1/3

Can someone explain how to get the answer of 81?

Use these two index laws in order:

ab÷ac=ab−ca^b \div a^c = a^{b-c}

(ab)c=ab×c(a^b)^c = a^{b \times c}
Reply 2
Avatar for ozmo19
ozmo19
OP
16
Original post by notnek
Use these two index laws in order:

ab÷ac=ab−ca^b \div a^c = a^{b-c}

(ab)c=ab×c(a^b)^c = a^{b \times c}


How do i go from (the cubed root of 9)^6 to 81 though? How can this be written out and explained?
Reply 3
Avatar for Notnek
Notnek
21
Original post by ozmo19
How do i go from (the cubed root of 9)^6 to 81 though? How can this be written out and explained?

What you've written is correct but I don't think you've followed the method properly.

(910÷94)13(9^{10} \div 9^4)^\frac{1}{3}

Start by using the first index law I mentioned to simplify the stuff inside the brackets:

910÷94=910−4=969^{10} \div 9^4 = 9^{10-4} = 9^6

So now you have

(910÷94)13=(96)13(9^{10} \div 9^4)^\frac{1}{3} = (9^6)^\frac{1}{3}

Can you see what to do next? If not, look back at the second index law I gave you.
Reply 4
Avatar for ozmo19
ozmo19
OP
16
Original post by notnek
What you've written is correct but I don't think you've followed the method properly.

(910÷94)13(9^{10} \div 9^4)^\frac{1}{3}

Start by using the first index law I mentioned to simplify the stuff inside the brackets:

910÷94=910−4=969^{10} \div 9^4 = 9^{10-4} = 9^6

So now you have

(910÷94)13=(96)13(9^{10} \div 9^4)^\frac{1}{3} = (9^6)^\frac{1}{3}

Can you see what to do next? If not, look back at the second index law I gave you.


Omg, im so stupid. It always stares me right in the face. 6*1/3=2 so 9^2=81.
Thanks; i know it was an easy question, haven't revised this section in a while:tongue:

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