why is it that ^4√112b^12 is written as 2b^3√7 instead of 16b^3√7?
I ask this question because doing problems that are a little bit more difficult are not making sense. with the above question we know that it's 2 times itself 4 times however we write it as 2b^3√7
However in this problem:
^3√320x^16y^12z^13
instead of it being 2 it's written as ^3√64. is it because there are only 2 sets of 2? If there were 3 sets of twos would it then be put outside of √?
I'm trying to make sure this is the reason. someone please help. Thanks.
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^4√112b^12
How about this approach:
^4√112b^12 = ^4√(112)*^4√(b^12)
This is because ^n√(a*b) = ^n√(a)*^n√(b)
Now notice that 112 = 7*16 = 7*2^4
^4√(112)*^4√(b^12) = ^4√(7*2^4)*^4√(b^12) = ^4√(2^4)*^4√(7)*^4√(b^12)
Now bear in mind that ^4√(2^4) = 2^(4/4) = 2^1 = 2
In other words, the fourth root of 2 elevated to four is 2.
Also, for this and for the rest of the problem, which you got right, it's good to know the following rule:
^n√(a^m) = a^(m/n)
So let's continue:
^4√(2^4)*^4√(7)*^4√(b^12) = 2*^4√(7)*b^(12/4) = 2*b^(12/4)*^4√(7) = 2b^3*^4√7
Notice that you forgot to specify that it is the fourth root of 7 in your answer.
Now let's use the same method for the second problem:
^3√320x^16y^12z^13 = ^3√(320)*^3√(x^16)*^3√(y^12)*^3√(z^13)
320 = 5*64 = 5*2^6
So:
^3√(320) = ^3√(5*2^6) = ^3√(5)*^3√(2^6) = ^3√(5)*2^(6/3) = ^3√(5)*2^2 = 4*^3√5
So the ^3√64 is wrong because 64 = 2^6 and this can be worked on, as I showed. Tell your teacher!
I hope this helped. I didn't do the rest of the second problem because it didn't seem to interest you.