(1/9)^(2/3)

[whysodifficult]

where have my sums gone wrong?

Expression 1: (1/9)^(2/3) = (9)^(3/2) = sqrt(9) ^ 3 = 3^ 3 = 27
Expression 2: (1/9)^(2/3) = cubic_root (1/81) = 1/3 or -1/3
Expression 3: (1/9)^(2/3) = (3^-2)^ (2/3) = 3^ (-4/3)

I know expression 3 is right. But why is expression 1 and 2 wrong?

 

[blamocur]

I know expression 3 is right. But why is expression 1 and 2 wrong?

Do you expect 3 different answers all to be right?

cubic_root (1/81) = 1/3 or -1/3

How do you get this?

(9)^(3/2) = sqrt(9) ^ 3 = 3^ 3 = 27

How do you change the exponent to compensate for inverting the base?

 

[fresh_42]

The rules are
$$\begin{array}{lll} \left(\dfrac{a}{b}\right)^{\frac{n}{m}}&=\left(\dfrac{a}{b}\right)^{n\cdot \frac{1}{m}}=\left(\left(\dfrac{a}{b}\right)^n\right)^{\frac{1}{m}}=\left(\dfrac{a^n}{b^n}\right)^{\frac{1}{m}}=\sqrt[m]{\dfrac{a^n}{b^n}}=\dfrac{\sqrt[m]{a^n}}{\sqrt[m]{b^n}} \end{array}$$In your case, we have $\sqrt[m]{b^n}=\sqrt[3]{3^4}=\sqrt[3]{81}$ but $81$ has four threes, not three threes. It is therefore no cubic root. All we can do is
$$\sqrt[3]{81}=\sqrt[3]{3^3\cdot 3}=\sqrt[3]{3^3}\cdot \sqrt[3]{3}=3\cdot\sqrt[3]{3}.$$

 

[khansaheb]

1/9)^(2/3) = (9)^(3/2)

Incorrect ........................ which law of mathematical operation you would use to get that (incorrect) equality?

(1/9)^(2/3) = cubic_root (1/81)

Incorrect ........................ which law of mathematical operation you would use to get that (incorrect) equality?

 

[Harry_the_cat]

Incorrect ........................ which law of mathematical operation you would use to get that (incorrect) equality?


Incorrect ........................ which law of mathematical operation you would use to get that (incorrect) equality?

(1/9)^(2/3) = cubic_root (1/81) is correct, but then whysodifficult takes the fourth root rather than the cube root.

 

[whysodifficult]

(1/9)^(2/3) = cubic_root (1/81) is correct, but then whysodifficult takes the fourth root rather than the cube root.

Yeah I realized what is wrong. 81 = 3^4 not 3^3. My bad.

 

[whysodifficult]

The rules are
$$\begin{array}{lll} \left(\dfrac{a}{b}\right)^{\frac{n}{m}}&=\left(\dfrac{a}{b}\right)^{n\cdot \frac{1}{m}}=\left(\left(\dfrac{a}{b}\right)^n\right)^{\frac{1}{m}}=\left(\dfrac{a^n}{b^n}\right)^{\frac{1}{m}}=\sqrt[m]{\dfrac{a^n}{b^n}}=\dfrac{\sqrt[m]{a^n}}{\sqrt[m]{b^n}} \end{array}$$In your case, we have $\sqrt[m]{b^n}=\sqrt[3]{3^4}=\sqrt[3]{81}$ but $81$ has four threes, not three threes. It is therefore no cubic root. All we can do is
$$\sqrt[3]{81}=\sqrt[3]{3^3\cdot 3}=\sqrt[3]{3^3}\cdot \sqrt[3]{3}=3\cdot\sqrt[3]{3}.$$

Sir pls tell me what is wrong with my mathematical identity:

(a/b)^n = (b/a)^-n = (b/a)^(1/n)

 

[khansaheb]

(1/9)^(2/3) = cubic_root (1/81) is correct, but then whysodifficult takes the fourth root rather than the cube root.

Correct you are!

My response should have quoted the next line indicating that

(1/3)^3 = 1/(27) ......NOT (1/81)

 

[Steven G]

where have my sums gone wrong?

Expression 1: (1/9)^(2/3) = (9)^(3/2) = sqrt(9) ^ 3 = 3^ 3 = 27
Expression 2: (1/9)^(2/3) = cubic_root (1/81) = 1/3 or -1/3
Expression 3: (1/9)^(2/3) = (3^-2)^ (2/3) = 3^ (-4/3)

I know expression 3 is right. But why is expression 1 and 2 wrong?

Expression 1: When you take the recriprocal of the base you do Not take the recriprocal of the power, rather you change the sign of the power.
Consider (1/2)^1/2, you would expect that result to be less than 1. Using your method, you would get 2² which is 4.

Expression 2: The result can't be -1/3 because if you cube it you'll get a negative number. Besides, the cube root of 81 is not 3 (or -3). The 4th root of 81 is +/- 3

 

[lookagain]

The 4th root of 81 is +/- 3

The 4th root of 81 is +3. It is the principal root.

 

[fresh_42]

Sir pls tell me what is wrong with my mathematical identity:

(a/b)^n = (b/a)^-n = (b/a)^(1/n)

$\left(\dfrac{a}{b}\right)^{-1}=\dfrac{b}{a}$ so your first equation $\left(\dfrac{a}{b}\right)^n=\left(\dfrac{b}{a}\right)^{-n}$ is correct.

The second is problematic (and wrong) $\dfrac{a^n}{b^n}=\dfrac{b^{-n}}{a^{-n}}=\left(\dfrac{b}{a}\right)^{-n}\neq \left(\dfrac{b}{a}\right)^{1/n}=\sqrt[n]{\dfrac{b}{a}} .$

There are several ways to see that.

Firstly, we have $something ^{-n} = something^{1/n}$ and one would expect, given both $something$ are equal, that this can only be if $-n=1/n,$ but that's wrong.

Second possibility: Set $n=2$ and $a=1.$ Then we have $b^{-2}=\dfrac{1}{b^2}$ on the left and $b^{1/2}=\sqrt{b}$ on the right. They are only equal if $b=1.$ So we set $b=2$ and see they are different.

Thirdly, we consider the definitions of $b^{-n} = \dfrac{1}{b^n}$ and $b^{1/n}=\sqrt[n]{b} .$ The former is the power of a reciprocal value, i.e. the solution of the equation $x\cdot b^n =1,$ and the latter is the n-th root, i.e. the solution of the equation $x^n=b.$ Those equations are very different, so their solution cannot be the same (except in some special cases, but not in general).