Join TSR now and get all your revision questions answeredSign up now

Proving an q^n is bigger or smaller than m^p Watch

    • Thread Starter
    Offline

    10
    ReputationRep:
    Hello

    Is there an efficient way of finding whichever one of 218 or 515 is bigger without a calculator, and not by multiplying until you get the value of the two?
    • Reporter Team
    Offline

    14
    ReputationRep:
    Instinct tells me to start from

    21^8 < 25^8

    But I'm almost certainly wrong.


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    Although I'm sure someone can come up with something better, I'll give it a shot.

    Attempt one (long):

    Spoiler:
    Show

    Lets say we have q^n and m^p

    To directly compare them, one easily method would to convert say, q, into the form m^y

    So to do this;

    q=m^y

    y=\frac{ln(q)}{ln(m)}

    Making; q^n \Rightarrow (m^{\frac{ln(p)}{ln(m)}})^n

    This can be expanded out using (a^n)^d = a^{nd}

    Thus,  q^n \equiv m^{\frac{nln(p)}{ln(m)}}

    Now we have them both as powers of m, so just compare the powers.

    In your example 21^8 and 5^{15}, if we use the same method, and put 21 as a power of 5.

    21=5^n

    n=\frac{ln21}{ln5}

    21^8 \equiv (5^{\frac{ln21}{ln5}})^8 \Rightarrow 5^{\frac{8ln21}{ln5}}

    \frac{8ln21}{ln5} \approx 15.1 &gt; 15

    21^8 &gt; 5^{15}



    Attempt 2 (same method, shorter). I was naive in not just considering this:

    Spoiler:
    Show


    q^n and m^p

    If nln(q)&gt;pln(m) \Rightarrow q^n&gt;m^p

    Or if:

    nln(q)&lt;pln(m) \Rightarrow q^n&lt;m^p



    Obviously this requires some rough calculation of logs, however I'm skeptical there is a method which involves next to no calculations. Considering the numbers could take any value, and just because the power of one is bigger, it doesn't necessarily mean the number is bigger.

    Edit- Found this: http://www.zachwg.org/logarithms.pdf, could be useful.
    Offline

    3
    ReputationRep:
    (Original post by Krollo)
    Instinct tells me to start from

    21^8 < 25^8

    But I'm almost certainly wrong.
    Posted from TSR Mobile
    lol
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by Phichi)
    x
    I can see where you are going with method one and I understand the basis of your workings, but when it comes to logarithms I clearly need to study it more (I'm in the AS year).

    Thanks for sharing, I will try and seek the knowledge you have demonstrated.
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by Krollo)
    Instinct tells me to start from

    21^8 < 25^8

    But I'm almost certainly wrong.


    Posted from TSR Mobile
    You are right that you are wrong but it's always good to try.
    I know a step I won't be trying now :P
    • Reporter Team
    Offline

    14
    ReputationRep:
    (Original post by Wunderbarr)
    You are right that you are wrong but it's always good to try.
    I know a step I won't be trying now :P
    Lol. For some reason my intuition often fails me for maths problems like this.


    Posted from TSR Mobile
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    (Original post by Wunderbarr)
    Hello

    Is there an efficient way of finding whichever one of 218 or 515 is bigger without a calculator, and not by multiplying until you get the value of the two?
    That's (21^2)^4 vs 125^5; 21^2 = 441 so we want 441^4 vs 125^5. The quotient is about 3.5^4 \times \frac{1}{125}, which if you can be bothered to find 3.5^4 which is roughly 150.1. Hence the quotient is greater than 1, so 125^5 &lt; 441^4; hence 21^8 is bigger.

    That only required finding 3.5^4 and approximating fairly cavalierly.

    Alternatively you can say \frac{21^8}{5^{15}} = 4.2^8 \times 5^{-7} which is 0.84^8 \times 5. You can work out 0.84^8 by repeated squaring if you want.

    Neither of those ways are nice, sadly.
 
 
 
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.