Turn on thread page Beta

Can you solve this without trial and error? watch

    • Thread Starter
    Offline

    10
    ReputationRep:
    5^x +4^x =8
    Online

    20
    ReputationRep:
    x must be less than 1
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by the bear)
    x must be less than 1
    I know, but can I figure the solution without trial and error?
    The exact value
    Online

    20
    ReputationRep:
    x is bigger than 0
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by mariejuana)
    I know, but can I figure the solution without trial and error?
    The exact value
    Change the bases.
    Offline

    3
    ReputationRep:
    Hay ho
    Offline

    3
    ReputationRep:
    On
    Offline

    3
    ReputationRep:
    Eb
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by RDKGames)
    Change the bases.
    I did and it gave me a weird answer.
    Offline

    3
    ReputationRep:
    Rty
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by RDKGames)
    What was the weird answer?
    i changed 4^x t0 2^2x. changed 8 to 2^3
    used logs to change 5 t0 a power of 2

    2^x= 5
    x=log5/log2
    so 2^log5/log2 x + 2^2x= 2^3
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by mariejuana)
    i changed 4^x t0 2^2x. changed 8 to 2^3
    used logs to change 5 t0 a power of 2

    2^x= 5
    x=log5/log2
    so 2^log5/log2 x + 2^2x= 2^3
    Sorry, ignore what I said as it doesn't apply here... long day

    You cannot find the exact solution to this equation because it is a transcendental equation. The best you can do is approximate the root by an iteration. One such iteration would be \displaystyle x_{n+1}=\log_5(8-4^{x_n}) and start with x_1=\frac{1}{2} as an example. This also means what you've done makes no sense.
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by RDKGames)
    Sorry, ignore what I said as it doesn't apply here... long day

    You cannot find the exact solution to this equation because it is a transcendental equation. The best you can do is approximate the root by an iteration. One such iteration would be x_{n+1}=\log_5(8-4^{x_n}) and start with x_1=\frac{1}{2} as an example. This also means what you've done makes no sense.

    Thank youuu!!
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by RDKGames)
    Sorry, ignore what I said as it doesn't apply here... long day

    You cannot find the exact solution to this equation because it is a transcendental equation. The best you can do is approximate the root by an iteration. One such iteration would be \displaystyle x_{n+1}=\log_5(8-4^{x_n}) and start with x_1=\frac{1}{2} as an example. This also means what you've done makes no sense.


    What about 5^x + 2*5^1-x =7 ?
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by mariejuana)
    What about 5^x + 2*5^1-x =7 ?
    You can easily solve that and find an exact solution.
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by RDKGames)
    You can easily solve that and find an exact solution.
    how ? xD
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by mariejuana)
    how ? xD
    Multiply through by 5^x to get (5^x)^2+10=7(5^x) which is a quadratic.
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by RDKGames)
    Multiply through by 5^x to get (5^x)^2+10=7(5^x) which is a quadratic.

    i don't understand it at allllllllllllll
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by mariejuana)
    i don't understand it at allllllllllllll
    Ok, it's literally nothing more difficult than solving a quadratic. Let y=5^x then you have y^2+10=7y which is the same as y^2-7y+10=0 which I'm sure you can solve before convering y back to 5^x
    • Thread Starter
    Offline

    10
    ReputationRep:
    (Original post by mariejuana)
    i don't understand it at allllllllllllll
    i do. doesn#t matter i was beingstupud
 
 
 
Poll
Cats or dogs?
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.