How do I solve $ n\int_R^{2R} dx/x = 2\int_{R/4}^R dx/x $?

Question

I was solving a physics problem when I got this result: $$ n\int_R^{2R} dx/x = 2\int_{R/4}^R dx/x $$. From this I'm supposed to get the result n=4. I tried solving this and got $$ n \ln (2R) - n \ln R = 2 \ln R - 2 \ln (R/4) $$. Is this correct? I am actually stuck over here. What should I be doing next?

Answer 1

Looks good so far, then recall logarithm properties to finish off: \begin{align}\require{cancel}n \ln 2 + \cancel{n \ln R} - \cancel{n \ln R} &= \cancel{2 \ln R} - \cancel{2 \ln R} + 2 \ln 4\\ n \ln 2& = 2 \ln 4 \\ &=4 \ln 2\\ \implies n &= 4.\end{align}

Answer 2

Use log rules:

$n [\ln (2) + ln (R) - \ln (R)] = 2[ \ln (R) - \ln (R) + ln (4)]$

$n=2 \times \frac {ln(4)}{ln(2)}=4$

Simples :)

Answer 3

Staring at a figure and doing some scalings in $x$- and $y$-direction it becomes clear that all integrals of the form $$\int_r^{2r}{dx\over x}\qquad(r>0)$$ have the same value. It is then obvious that $n=4$.

Apart from this: You have to use that $\log(uv)=\log u+\log v$.

Answer 4

As you said: $$n\int_{R}^{2R}{\frac{dx}{x}}=n(\ln(2R)-\ln(R))$$ so this is correct. $$2\int_{\frac{R}{4}}^{R}{\frac{dx}{x}}=2(\ln(R)-\ln(\frac{R}{4}))$$ and since the two integrals are equal, $$=n(\ln(2R)-\ln(R))=2(\ln(R)-\ln(\frac{R}{4}))$$ now note that: $$2(ln(R)-ln(\frac{R}{4}))=2\ln(4)$$ and: $$n(ln(2R)-ln(R))=n\ln(2)$$ so: $$n\ln(2)=2\ln(4)$$ since: $2\ln(4)=4\ln(2)$

so: $n\ln(2)=4\ln(2)$ $$\therefore n=4$$ Thats what I got

Answer 5

Hint

Your calculation is correct. If you use the formula \begin{align} \log a - \log b = \log a/b, \end{align} you will get the answer $n = 4$.