# Simplifying natural logs

Watch
Announcements
#1
I hope everyone is doing good today. I understand how to do this question and all the steps involved in it. But however I don’t get how to simplify it like they did here. (If I want to click on explain more I have to pay).
Last edited by Eris13696; 1 month ago
0
1 month ago
#2
(Original post by Eris13696)
I hope everyone is doing good today. I understand how to do this question and all the steps involved in it. But however I don’t get how to simplify it like they did here. (If I want to click on explain more I have to pay).
which bit is confusing you - these are just standard log rules so

2ln(3) = ln(3^2) = ln 9
2xln3 = ln(3^(2x)) etc
0
#3
(Original post by davros)
which bit is confusing you - these are just standard log rules so

2ln(3) = ln(3^2) = ln 9
2xln3 = ln(3^(2x)) etc
What’s confusing me is how the simplify it, this is how I do it. This is the answer I get without using their simplification. If you put values in both mine and theirs, you get the same thing. If I kept this in the exam, will it be fine or do I have to always simplify?
Last edited by Eris13696; 1 month ago
0
1 month ago
#4
(Original post by Eris13696)
What’s confusing me is how the simplify it, this is how I do it. This is the answer I get without using their simplification. If you put values in both mine and theirs, you get the same thing. If I kept this in the exam, will it be fine or do I have to always simplify?
Well, your answer isn't technically wrong (as far as I can see from a quick glance) but you've made an awful lot of extra work for yourself!

is just of the form with k = -2ln3, so you should be able to integrate this straight off to get:

as it is a standard integral.

You can then choose whether to write the solution as a power of e (which is how the original integrand looks), or convert it to a power of 3 or a power of 9. Personally I don't see anything to choose between and
0
#5
(Original post by davros)
Well, your answer isn't technically wrong (as far as I can see from a quick glance) but you've made an awful lot of extra work for yourself!

is just of the form with k = -2ln3, so you should be able to integrate this straight off to get:

as it is a standard integral.

You can then choose whether to write the solution as a power of e (which is how the original integrand looks), or convert it to a power of 3 or a power of 9. Personally I don't see anything to choose between and
I’m so grateful that you answered me, you’re the best one at simplifying math here 😭 I spent 2 hours yesterday trying to understand how it’s done and then I found these steps on the app so I just went with whatever I had. Your method is done in like 2 steps compared to the one I used, thank you <3
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### How would you feel if uni students needed to be double vaccinated to start in Autumn?

I'd feel reassured about my own health (46)
15.03%
I'd feel reassured my learning may be less disrupted by isolations/lockdowns (98)
32.03%
I'd feel less anxious about being around large groups (38)
12.42%
I don't mind if others are vaccinated or not (26)
8.5%
I'm concerned it may disadvantage some students (17)
5.56%
I think it's an unfair expectation (78)
25.49%
Something else (tell us in the thread) (3)
0.98%