How do you condense #2ln6 - ln3#?

1 Answer
May 24, 2016

Answer:

#ln(12)#

Explanation:

Logs are subtracted if the source numbers are divided.

If the source number is raised to a power than you can multiply the loge by the value of that power. Suppose you had #log(x^a)# then this could be written as #alog(x)#

So #2ln(6)-ln(3)" " ->" " ln(6^2)-ln(3)" " ->" " ln(6^2/3)#

#=ln(12)#