Join TSR now and get all your revision questions answeredSign up now
    • Thread Starter
    Offline

    5
    ReputationRep:
    Solve for x:



    Where do I even start? :s Bases are different.

    I tried exponentiating both sides (raising both sides to the power of x and then also tried the same but to the power of 8, also tried expressing the whole expression in terms of e, rather than 8. Got somewhere but nowhere useful, I believe.

    Thanks!

    Sid
    • Study Helper
    Offline

    13
    (Original post by Sidhant Shivram)
    ...
    Can't see the image.

    Guessing - you're going to need to change the base.
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    (Original post by Sidhant Shivram)
    Solve for x:



    Where do I even start? :s Bases are different.

    I tried exponentiating both sides (raising both sides to the power of x and then also tried the same but to the power of 8, also tried expressing the whole expression in terms of e, rather than 8. Got somewhere but nowhere useful, I believe.

    Thanks!

    Sid
    As ghostwalker said, the picture isn't there. However, it may be useful to remember that \log(a^b) = b \log(a).
    • Thread Starter
    Offline

    5
    ReputationRep:
    (Original post by ghostwalker)
    Can't see the image.

    Guessing - you're going to need to change the base.
    (Original post by Smaug123)
    As ghostwalker said, the picture isn't there. However, it may be useful to remember that \log(a^b) = b \log(a).


    18.b) The 2nd part
    Attached Images
      
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    (Original post by Sidhant Shivram)


    18.b) The 2nd part
    \log_x(8) = \frac{\log_e(8)}{\log_e(x)} = \log_e(8^{1/\log_e(8)}) unless x \leq 0 or x=1. That's how I'd end up doing it, I think.
    • Thread Starter
    Offline

    5
    ReputationRep:
    (Original post by Smaug123)
    \log_x(8) = \frac{\log_e(8)}{\log_e(x)} = \log_e(8^{1/\log_e(8)}) unless x \leq 0 or x=1. That's how I'd end up doing it, I think.
    Name:  ImageUploadedByTapatalk1419699803.988034.jpg
Views: 65
Size:  263.2 KB

    Ah yes. Makes sense. I used your method. Got some sensible answers. However, the answers given are : 4 and 16 x root2 :/


    Sent from my iPhone using Tapatalk
    Offline

    15
    ReputationRep:
    you can write the 2 on its own as 2log33
    Offline

    3
    ReputationRep:
    (Original post by the bear)
    you can write the 2 on its own as 2log33
    Exactly the same as what I was thinking.
    May be better to write it as the logarithm of 9 to base 3 however.
    • Thread Starter
    Offline

    5
    ReputationRep:
    (Original post by the bear)
    you can write the 2 on its own as 2log33
    (Original post by MathMeister)
    Exactly the same as what I was thinking.
    May be better to write it as the logarithm of 9 to base 3 however.
    The first part is fine. its the 2nd part, part (b), i.e., thats bothering me.
    Offline

    3
    ReputationRep:
    (Original post by Sidhant Shivram)
    The first part is fine. its the 2nd part, part (b), i.e., thats bothering me.
    Change of base rule.
    Offline

    15
    ReputationRep:
    (Original post by Smaug123)
    \log_x(8) = \frac{\log_8(8)}{\log_8(x)} .
    this is easier
    • Thread Starter
    Offline

    5
    ReputationRep:
    (Original post by MathMeister)
    Change of base rule.
    (Original post by the bear)
    this is easier
    attachment.php ?

    Got some sensible answers. However, the answers given are : 4 and 16 x root2 :/
    Offline

    3
    ReputationRep:
    (Original post by Sidhant Shivram)
    attachment.php ?

    Got some sensible answers. However, the answers given are : 4 and 16 x root2 :/
    Haven't got a pen or pencil (or paper) on me atm but by just looking at the Q-maybe use the change of base rule and try to form a quadratic? (remembering the logarithmic function is defined for positive values of x.
    I may be wrong however as I have not tried it out myself.
    Offline

    1
    ReputationRep:
    (Original post by Sidhant Shivram)
    ...
    The answers given check out.

    Your fourth line of working does not follow from third.
    • Thread Starter
    Offline

    5
    ReputationRep:
    (Original post by lazy_fish)
    The answers given check out.

    Your fourth line of working does not follow from third.
    Oops! Got it! thank you so much!
 
 
 
Poll
If you won £30,000, which of these would you spend it on?
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

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