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Comment on Cube Root of a Decimal
Way simpler than thought it
((0.000008)^1/3 )^-1 = 1/((0
= 1/(( (0.02)^3 )^1/3) = 1/ ((0.02)^3/3) = 1/0.02
= 1/(2/100) = 100/2 = 50
How do we know the cube root
Good question.
Good question.
We know that if ∛x = y, then x = y³ = (y)(y)(y)
For example, if ∛8 = 2, then 8 = 2³ = (2)(2)(2)
Likewise, if ∛(1,000,000) = y, then 1,000,000 = y³
= (y)(y)(y)
Now notice that, whenever we multiply powers of 10, we add the number of zeros.
For example, (100)(100,000) = 10,000,000
In other words, (power of 10 with TWO zeros)(power of 10 with FIVE zeros) = power of 10 with SEVEN zeros
Now notice that 1,000,000 has SIX zeros.
So, if we place TWO zeros in each set of brackets (above), we get:
(100)(100)(100), which equals 1,000,000
So, ∛(1,000,000) = 100
Does that help?
Cheers,
Brent
I like this method but
1. 1,000,000 = 10^6
If STUCK about how to write one million as a power of 10, think of an easy example:
100 = 10^2 and
100 = a ONE followed by *2* zeros
so
1,000,000 = a ONE followed by *6* zeros = (10^6)
2. ∛k = k^(1/3), [*see below], so
3. ∛(10^6) = (10^6)^[1/3]
4. In step 3, we have a power to a power, so multiply the powers
10^[6 x (1/3)] =
10^(6/3) =
10^(2) = 100
So the cube root of
1,000,000 = 100
(In other words, use a fractional exponent to represent the root, then multiply root times power. The answer is the value you need.)
*I can't type Brent's cool little graphic about the property that
▪︎the nth root equals a fractional power (1/n)▪︎
but you can see that graphic and how it works if you watch here:
https://www.greenlighttestprep.com/module/gre-powers-and-roots/video/1038
(esp 3:12 to end)
Study Guide, Step 8
POWERS AND ROOTS MODULE
Lesson: Other Roots
(I find it easier to calculate not-obvious roots this way.)
How do we know to re-write it
We can work our way to that
We can work our way to that as follows:
0.8 = 8/10
0.08 = 8/100
0.008 = 8/1,000
0.0008 = 8/10,000
0.00008 = 8/100,000
0.000008 = 8/1,000,000
Cheers,
Brent
Is there an alternative way
You bet here's another
You bet here's another solution:
We'll use the fact that ∛(ab) = (∛a)(∛b)
We'll calculate: [∛(0.000008)]^(-1)
Let's first deal with: ∛(0.000008)
Rewrite as: ∛(8 x 0.000001)
Rewrite as: ∛(8 x 10^-6)
Rewrite as: (∛8)(∛10^-6)
Evaluate: (2)(10^-2)
Now we're deal with the ^(-1) part
We now have: [(2)(10^-2)]^(-1)
Apply Power of a Product law: [2^(-1)][(10^-2)^(-1)]
Simplify: [2^(-1)][10^2]
Simplify: [1/2][100]
Simplify: 50
Cheers,
Brent