Law of Exponents

[mdavis]

Could someone explain, step by step, how to solve this problem?

(1/16)^-1/4

[dmateer25]

Because it is raised to the -1/4 you need to take the inverse of 1/16.

So it will become 16^1/4

Now from here you take 16 to the power in the numerator.

16^1 = 16

and then because the denominator is 4 you take the 4th root of 16

the 4th root of 16 = 2

[DanaJ]

Let's take everything step by step, then.

You have (1/16) raised to the power -1/4. We should tackle the minus first. That minus over there basically tells us that you should "flip" the fraction, turning 1/16 into 16/1 or 16. This is what a negative power means: just tun it over!
The general rule is that (a/b)^(-n) = (b/a)^n.

Now, let's see what we get: (1/16)^(-1/4) = 16^(1/4). Now, this 1/4 actually stands for the fourth degree root of 16, which will be 2, since 2^4 = 16. Following this rule, we get that sqrt(n) = n^(1/2). It's just another notation for roots.
The general rule over here is: a^(1/n) = the n-th degree root of a.
Let's take some more examples:
81^(1/4) = 3 (the fourth degree root), since 3^4 = 81
8^(1/3) = 2 (the third degree root), since 2^3 =8.
64^(1/6) = 2 (the sixth degree root), since 2^6 = 64.

[vladmire]

Let's take everything step by step, then.

You have (1/16) raised to the power -1/4. We should tackle the minus first. That minus over there basically tells us that you should "flip" the fraction, turning 1/16 into 16/1 or 16. This is what a negative power means: just tun it over!
The general rule is that (a/b)^(-n) = (b/a)^n.

Now, let's see what we get: (1/16)^(-1/4) = 16^(1/4). Now, this 1/4 actually stands for the fourth degree root of 16, which will be 2, since 2^4 = 16. Following this rule, we get that sqrt(n) = n^(1/2). It's just another notation for roots.
The general rule over here is: a^(1/n) = the n-th degree root of a.
Let's take some more examples:
81^(1/4) = 3 (the fourth degree root), since 3^4 = 81
8^(1/3) = 2 (the third degree root), since 2^3 =8.
64^(1/6) = 2 (the sixth degree root), since 2^6 = 64.

What if the problem was (1/16)^-3/4 what would I do with the 3

[dmateer25]

Let's take everything step by step, then.

You have (1/16) raised to the power -1/4. We should tackle the minus first. That minus over there basically tells us that you should "flip" the fraction, turning 1/16 into 16/1 or 16. This is what a negative power means: just tun it over!
The general rule is that (a/b)^(-n) = (b/a)^n.

Now, let's see what we get: (1/16)^(-1/4) = 16^(1/4). Now, this 1/4 actually stands for the fourth degree root of 16, which will be 2, since 2^4 = 16. Following this rule, we get that sqrt(n) = n^(1/2). It's just another notation for roots.
The general rule over here is: a^(1/n) = the n-th degree root of a.
Let's take some more examples:
81^(1/4) = 3 (the fourth degree root), since 3^4 = 81
8^(1/3) = 2 (the third degree root), since 2^3 =8.
64^(1/6) = 2 (the sixth degree root), since 2^6 = 64.

What if the problem was (1/16)^-3/4 what would I do with the 3

(1/16)^-3/4

because the 3 is negative you still take the inverse.

So it would become 16^3/4.

Now what you do is take the base 16 to 3rd power. and then you take the 4th root of this.

So it would be the 4th root of 16^3 which would equal 8.

[chintudave]

(1/16)^-1/4

= 16 ^ 1/4

i.e. 4th root of 16.

As others have said the answer is 4... however if this question shows up in DS... the options are 2, -2, 2i & -2i... so there is no single value to this problem.