How does this work?

[Adi]

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?

 

[MarkFL]

Yes, that's not the result of good algebra. The first equation implies:

[MATH]3=\frac{k}{k^{2/3}}=k^{1/3}[/MATH]

 

[Adi]

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

 

[Adi]

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

 

[Harry_the_cat]

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]

The first equation says:

[MATH]k^{2/3}=\frac{1}{3}k[/MATH]
Now, suppose we multiply both sides by 3...we get:

[MATH]3k^{2/3}=k[/MATH]
Now, divide both sides by $k^{2/3}$

[MATH]3=\frac{k}{k^{2/3}}[/MATH]
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]

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:
$\displaystyle \frac{k}{k^{\frac{2}{3}}}=3$
$\displaystyle k^{1-\frac{2}{3}} =3$
$\displaystyle k^{\frac{1}{3}} = 3$
$\displaystyle k = 3^3$
$\displaystyle k=27$