Express as a simple logarithm 2ln6 - ln3 .

We start with: 2ln6 - ln3 ... First, we rewrite this expression as: ln6 + ln6 - ln3 ... Next, we rewrite this as: ln(23) + ln(23) - ln3 ... Using the log rule logaxy = logax + logxy, we express this as ln2 + ln3 + ln2 + ln3 - ln3 ... We simplify this to ln2 + ln2 + ln3 ... Using the log rule logax + logay = logaxy, we express this as ln (223) ... Finally, we can simplify this to ln12. Alternative method: We start with: 2ln6 - ln3 ... First, using the log rule: ylogax = loga(xy) we express this as ln(62)- ln3 ... Next, we rewrite this as: ln36 - ln3 ... Using the log rule logax - logxy = loga(x/y) we express this as ln(36/3) ... Finally, we can simplify this to ln12

Answered by Maths tutor

5134 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do you differentiate parametric equations?


Integrate using by parts twice : ∫e^(x)*(cos(x))dx


Given that y=4x^3-(5/x^2) what is dy/dx in it's simplest form?


Given x = 3sin(y/2), find dy/dx in terms of x, simplifying your answer.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning