# How does this work?

##### New member
Hey guys pretty simple algebra question. I don't understand how this step works in the solution to an economics problem I'm doing. In particular what operation has been done? Has it been divided by k on both sides? If yes then how does 1/3k/k=3?

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#### MarkFL

##### Super Moderator
Staff member
Yes, that's not the result of good algebra. The first equation implies:

$$\displaystyle 3=\frac{k}{k^{2/3}}=k^{1/3}$$

##### New member
I'm sorry I still don't understand. Can you explain in a more for dummies way?

##### New member
I'm sorry I still don't understand. Can you explain in a more for dummies way?
Here is the full solution for your reference

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#### Harry_the_cat

##### Senior Member
Hey guys pretty simple algebra question. I don't understand how this step works in the solution to an economics problem I'm doing. In particular what operation has been done? Has it been divided by k on both sides? If yes then how does 1/3k/k=3?
The second line is incorrect. It should read $$\displaystyle \frac {k^{\frac{2}{3}}}{k} = \frac{1}{3}$$ which is equivalent to what MarkFL said

#### MarkFL

##### Super Moderator
Staff member
The first equation says:

$$\displaystyle k^{2/3}=\frac{1}{3}k$$

Now, suppose we multiply both sides by 3...we get:

$$\displaystyle 3k^{2/3}=k$$

Now, divide both sides by $$k^{2/3}$$

$$\displaystyle 3=\frac{k}{k^{2/3}}$$

In the second image you posted, they do eventually get there, but they introduce other algebraic errors that magically cancel out the first. That solution is very poorly written.

#### Harry_the_cat

##### Senior Member
There are 2 errors in the full "solution" you have supplied which effectively cancel each other out. The first error is from line 1 to 2 and the second is from line 3 to 4.
Continuing from my correct second line in my post above:
Reciprocating gives:
$$\displaystyle \frac{k}{k^{\frac{2}{3}}}=3$$
$$\displaystyle k^{1-\frac{2}{3}} =3$$
$$\displaystyle k^{\frac{1}{3}} = 3$$
$$\displaystyle k = 3^3$$
$$\displaystyle k=27$$