Asked by Eve
I am trying to figure out how to evaluate 100^-3/2. I'm going by the example (64)^2/3=16. IN the example, 64^2/3 is changed into 64^1/3^2=4^2=16. But I don't get why out of all numbers it is changed into 4^2. I know 16 goes into 64 four times, but why is the answer 16 and what becomes of the 1/3? So how would this apply to 11^-3/2? Thanks for your help!
Answers
Answered by
MathMate
First, you'd have to adhere to the rules of priority.
The expression 64<sup>2/3</sup> is much clearer.
The 2 in the exponent requires us to square the base, and the 3 in the denominator represents (1/3) means we need to take the cube root of the base.
So whether you do ∛((64)²) or (∛64)³ will give the same answer, i.e. ∛(4096)=16 or (4)²=16. The 4 comes from the cube root of 64, or 64<sup>1/3</sup>.
For 11<sup>-3/2</sup> it is the same procedure, by following the laws of exponents:
x<sup>a+b</sup> = x<sup>a</sup> × x<sup>b</sup>
x<sup>ab</sup> = (x<sup>a</sup>)<sup>b</sup>
x<sup>-a</sup> = 1/x<sup>a</sup>
x<sup>1/a</sup> = ath root of x.
Therefore
11<sup>-3/2</sup>
=1/11<sup>3/2</sup>
=1/∛(11³)
=1/∛(1331)
The expression 64<sup>2/3</sup> is much clearer.
The 2 in the exponent requires us to square the base, and the 3 in the denominator represents (1/3) means we need to take the cube root of the base.
So whether you do ∛((64)²) or (∛64)³ will give the same answer, i.e. ∛(4096)=16 or (4)²=16. The 4 comes from the cube root of 64, or 64<sup>1/3</sup>.
For 11<sup>-3/2</sup> it is the same procedure, by following the laws of exponents:
x<sup>a+b</sup> = x<sup>a</sup> × x<sup>b</sup>
x<sup>ab</sup> = (x<sup>a</sup>)<sup>b</sup>
x<sup>-a</sup> = 1/x<sup>a</sup>
x<sup>1/a</sup> = ath root of x.
Therefore
11<sup>-3/2</sup>
=1/11<sup>3/2</sup>
=1/∛(11³)
=1/∛(1331)
Answered by
MathMate
11<sup>-3/2</sup>
=1/11<sup>3/2</sup>
=1/√(11³)
=1/√(1331)
=1/11<sup>3/2</sup>
=1/√(11³)
=1/√(1331)