Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    2
    ReputationRep:
    When I get an equation of 'f(x)= ln(2+3x)', how do I derive it from the original equation of 'ln x'.

    Am I right in thinking that to get from 'ln x' to 'f(x)= ln(2+3x)', I divide the x values of 'ln x' by three, and then shift the whole curve 2 units to the left ? Also, where would the asymtote be in 'f(x)= ln(2+3x)'?



    my other question is, when find I was finding the inverse of 'ln(5x-2)', I got: 'e^(x+5) / 5'. But in the answers, it has exponential just to the power of 'x', whereas I thought it would be to the power of 'x+5'. Can someone explain this for me?
    Offline

    1
    ReputationRep:
    (Original post by W.H.T)
    When I get an equation of 'f(x)= ln(2+3x)', how do I derive it from the original equation of 'ln x'.

    Am I right in thinking that to get from 'ln x' to 'f(x)= ln(2+3x)', I divide the x values of 'ln x' by three, and then shift the whole curve 2 units to the left ? Also, where would the asymtote be in 'f(x)= ln(2+3x)'?



    my other question is, when find I was finding the inverse of 'ln(5x-2)', I got: 'e^(x+5) / 5'. But in the answers, it has exponential just to the power of 'x', whereas I thought it would be to the power of 'x+5'. Can someone explain this for me?
    Yes, remember, stretches and reflections first, translations last. The asymptote would be when x = -2/3 because that makes the equation ln0 which is undefined.

    For the inverse of ln(5x-2)
    The steps are simple so I might as well just show you:

    y = ln(5x-2)
    x = ln(5y-2)
    e^x = 5y - 2

    5y = e^x + 2
    y = (e^x +2)/5
    Offline

    2
    ReputationRep:
    1. yup u r right, the asymtote will be at -2/3 i think

    2. I got the inverse as (e^x + 2)/5 if its right than I can explain
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.