0
$\begingroup$

Goal is to find both $\beta$ and $\omega$. I already have the answer here, but I'm confused as to how to get it.

$\log_6 250 - \log_\beta 2 = 3 \log_\beta \omega$

This is what I did:

$\log_6 250 = \log_\beta \omega^3 + \log_\beta 2$

$\log_6 250 = \log_\beta 2 \omega^3$

$\frac{\log 250}{\log 6} = \frac{\log 2\omega^3}{\log \beta}$

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:

$\frac {\log 64}{\log 4} = \frac{\log 27}{\log 3}$

$64 \ne 27, 4 \ne 3$

How would one solve this problem, and is the path I went correct?

$\endgroup$
2
$\begingroup$

From:

$\log_{\beta} (2w^{3}) = \log_{6}(250)$, we can set $\beta = 6$ since we simply want to find a solution and not all solutions.

Equating $2w^3=250$ lets us solve for a unique $w$.

$\endgroup$
  • $\begingroup$ Oh. So simple in hindsight, and was just overthinking it. $\endgroup$ – user154989 Dec 10 '15 at 15:59
1
$\begingroup$

HINT:

$$\log_6(250)-\log_\beta(2)=3\log_\beta(\omega)\Longleftrightarrow$$ $$\frac{\ln(250)}{\ln(6)}-\frac{\ln(2)}{\ln(\beta)}=\frac{3\ln(\omega)}{\ln(\beta)}\Longleftrightarrow$$ $$\frac{\ln(\beta)\ln(250)}{\ln(\beta)\ln(6)}-\frac{\ln(6)\ln(2)}{\ln(6)\ln(\beta)}=\frac{3\ln(\omega)}{\ln(\beta)}\Longleftrightarrow$$ $$\frac{\ln(\beta)\ln(250)-\ln(6)\ln(2)}{\ln(\beta)\ln(6)}=\frac{3\ln(\omega)}{\ln(\beta)}\Longleftrightarrow$$ $$\ln(\beta)\left(\ln(\beta)\ln(250)-\ln(6)\ln(2)\right)=3\ln(\beta)\ln(6)\ln(\omega)\Longleftrightarrow$$ $$3\ln(6)\ln(\beta)\ln(\omega)=\ln(250)\ln^2(\beta)-\ln(2)\ln(6)\ln(\beta)$$

$\endgroup$
0
$\begingroup$

Formula of changing bases is $\log_b x=\frac{\log_a x}{\log_a b}$. From this we get $\log_{\beta}(2w^3)^{\log_{\beta} 6}=\log_{\beta}250$ which implies (log is 1-to-1) $$(2w^3)^{\log_{\beta} 6}=250=2\cdot5^3$$ which admits the solution $\log_{\beta} 6=1$ with $\beta =6$ and $\omega=5$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.