Solving a log equation for two variables Goal is to find both beta and omega . I already have the answer here, but I'm confused as to how to get it. log_6 250 - log_beta 2 = 3 log_beta omega

Solving a log equation for two variables
Goal is to find both $\beta$ and $\omega$. I already have the answer here, but I'm confused as to how to get it.
${\mathrm{log}}_{6}250-{\mathrm{log}}_{\beta }2=3{\mathrm{log}}_{\beta }\omega$
This is what I did:
${\mathrm{log}}_{6}250={\mathrm{log}}_{\beta }{\omega }^{3}+{\mathrm{log}}_{\beta }2$
${\mathrm{log}}_{6}250={\mathrm{log}}_{\beta }2{\omega }^{3}$
$\frac{\mathrm{log}250}{\mathrm{log}6}=\frac{\mathrm{log}2{\omega }^{3}}{\mathrm{log}\beta }$
And I am stuck here. The answer states that $\omega =5$ and $\beta =6$, which after entering it, is correct, but I don't know how it got to that point.
Looking directly at the last part I ended up with, it seems like you're supposed to equate the tops to one another and bottoms to one another, which would get you the answer, but with another example, that clearly does not work:
$\frac{\mathrm{log}64}{\mathrm{log}4}=\frac{\mathrm{log}27}{\mathrm{log}3}$
$64\ne 27,4\ne 3$
How would one solve this problem, and is the path I went correct?
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periasemdy
From:
${\mathrm{log}}_{\beta }\left(2{w}^{3}\right)={\mathrm{log}}_{6}\left(250\right)$, we can set $\beta =6$ since we simply want to find a solution and not all solutions.
Equating $2{w}^{3}=250$ lets us solve for a unique $w$.