How can logarithmic expressions be simplified effectively?

[yungman]

I want to verify this:

2ln(x)-ln(2x)=ln(x^2)-ln(2x)=ln\left(\frac{x^2}{2x}\right)=ln\left(\frac x 2\right)
ln(2x)-ln(x)=\ln\left(\frac {2x}{x}\right)=ln(2)

Thanks

 

[phosgene]

They're pretty straight applications of the log laws

log_{c}(a^{b}) = blog_{c}(a)

log_{c}(a) - log_{c}(b) = log_{c}(\frac{a}{b})

 

[SammyS]

I want to verify this:

2ln(x)-ln(2x)=ln(x^2)-ln(2x)=ln\left(\frac{x^2}{2x}\right)=ln\left(\frac x 2\right)
ln(2x)-ln(x)=\ln\left(\frac {2x}{x}\right)=ln(2)

Thanks

You can also do the first one as

$\displaystyle 2\ln(x)-\ln(2x)=2\ln(x)-(\ln(x)+\ln(2))=\ln(x)-\ln(2)=\ln(x/2)$

 

[yungman]

You can also do the first one as

$\displaystyle 2\ln(x)-\ln(2x)=2\ln(x)-(\ln(x)+\ln(2))=\ln(x)-\ln(2)=\ln(x/2)$

Thanks, I am so rusty on these math as I don't use it often! I was stuck for a day in the other thread about sine and cosine integrals because of this. All of a sudden, I remember all the log things and it answer my question there.

Many thanks.