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    could anyone tell me how to solve logs like these

    Find x

    1.) 2^x+1 = 3^3x-2

    2.) 5^2x-3 = 7^3x-1

    I don't have a clue :P
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    (Original post by jacksonmeg)
    could anyone tell me how to solve logs like these

    Find x

    1.) 2^x+1 = 3^3x-2

    2.) 5^2x-3 = 7^3x-1

    I don't have a clue :P
    Is the first question 2^{x+1} = 3^{3x-2}
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    yes
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    Can you put that in brackets please.
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    I thought this thread was entitled "legs." What am I like?
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    Take logs of both sides to the base of either the 3 or 2 for the first one so the log cancels out on one of the sides. Then just work from there.


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    (Original post by jacksonmeg)
    yes
    Are you aware that a^x = b^y can be re-written as y log a = x log b
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    Yes, but I don't know how to solve them, need help urgently, can't find any worked examples like these
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    (Original post by jacksonmeg)
    Yes, but I don't know how to solve them, need help urgentl
    What have you done so far?
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    (X+1)log2 - (3x-2)log3 = 0 :/

    I know when you minus you divide them but I got the wrong answer :/
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    (Original post by jacksonmeg)
    could anyone tell me how to solve logs like these

    Find x

    1.) 2^x+1 = 3^3x-2

    2.) 5^2x-3 = 7^3x-1

    I don't have a clue :P

    You need to take logs of both sides (Any base, but either 2 or 3 would be best). Then one you have that, use log laws to take out the powers:
    x+1 = (3x-2)log(2,3)
    x+1 = 3x*log(2,3) -2log(2,3)
    x+3x*log(2,3) = -1-2log(2,3)
    x(1+3*log(2,3)) = -1-2log(2,3)
    x = (-1-2log(2,3))/(1+3log(2,3))

    There we go
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    (Original post by Dingusnin)
    You need to take logs of both sides (Any base, but either 2 or 3 would be best). Then one you have that, use log laws to take out the powers:
    x+1 = (3x-2)log(2,3)
    x+1 = 3x*log(2,3) -2log(2,3)
    x+3x*log(2,3) = -1-2log(2,3)
    x(1+3*log(2,3)) = -1-2log(2,3)
    x = (-1-2log(2,3))/(1+3log(2,3))

    There we go
    Perhaps spoiler your solution so the OP can have a go.
    Full solutions should only be provided if absolutely necessary.
    Please Read.
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    (Original post by Dingusnin)
    You need to take logs of both sides (Any base, but either 2 or 3 would be best). Then one you have that, use log laws to take out the powers:
    x+1 = (3x-2)log(2,3)
    x+1 = 3x*log(2,3) -2log(2,3)
    x+3x*log(2,3) = -1-2log(2,3)
    x(1+3*log(2,3)) = -1-2log(2,3)
    x = (-1-2log(2,3))/(1+3log(2,3))

    There we go
    i don't get what you've done :/
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    (Original post by jacksonmeg)
    i don't get what you've done :/
    Take logs of both sides.
    E.g \ log_2
    Spoiler:
    Show
    \ 2^{x+1} = 3^{3x-2} \Rightarrow x+1 = (3x-2)log_23

    How do you think we should proceed?
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    Sorry, i will spoiler-it next time

    what i have done is that i have taken logs (of base 2) on either side, and rearranged in order to solve for x.
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    What's the comma mean? Logs can go die, going to fail C2
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    (Original post by jacksonmeg)
    What's the comma mean? Logs can go die, going to fail C2
    Sorry, i should have posted using LaTeX, What i meant by that is:
    log(base,x)
    in other words: log(2,8) = 3, Because 2^3 = 8
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    (Original post by joostan)
    Take logs of both sides.
    E.g \ log_2
    Spoiler:
    Show
    \ 2^{x+1} = 3^{3x-2} \Rightarrow x+1 = (3x-2)log_23

    How do you think we should proceed?
    I don't have a clue :/
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    (Original post by jacksonmeg)
    I don't have a clue :/
    Well how would you go about isolating x?
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    Idk :/ none of the rules I've got seem to work
 
 
 
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