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    Hi, im working on the following question, but i havent been taught to use log when solving this. It asks you to solve it for x. Wondered if someone could remind me how to do it without using log??

    5^(2x)=1/125

    Again, haven't been taught to use log, taught another way. Any help??
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    (Original post by LuckyEms)
    Hi, im working on the following question, but i havent been taught to use log when solving this. It asks you to solve it for x. Wondered if someone could remind me how to do it without using log??

    5^(2x)=1/125

    Again, haven't been taught to use log, taught another way. Any help??
    logs aren't needed here, 5^x=1/125, now what power of x would give you 1/125?
    remember x^-^1 would give you 1/x
    then once you found the power make it equal to 2x then you can rearrange to find what x is from there.
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    Take loagarithms of both sides, remembering that log(A^B) = B*log(A). There is an easier way, however - write 1/125 as a power of five and then equate this to 2x.
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    5^3 = 125, so 5^-3 = 1/125, then you just need to divide -3 by 2 to give x on its own (-1.5)
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    (Original post by jadecross)
    logs aren't needed here, 5^x=1/125, now what power of x would give you 1/125?
    remember x^-^1 would give you 1/x
    5^-3 gives 1/125
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    (Original post by LuckyEms)
    5^-3 gives 1/125
    -3=2x now solve x.
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    (Original post by jadecross)
    -3=2x now solve x.
    Ahh thank you
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    (Original post by LuckyEms)
    Ahh thank you
    you're welcome.
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    (Original post by JackS94)
    5^3 = 125, so 5^-3 = 1/125, then you just need to divide -3 by 2 to give x on its own (-1.5)
    don't post solutions, read the rules.
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    Thank you all!!

    Mind someone helping again?

    The next one is 9^x+1 = 3^x sqrt(27).

    I started by doing 3^x x 27^1/2 to get 81^x+1/2
    So i had 9^x+1=81^x+1/2

    Am i going in the right direction??
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    (Original post by jadecross)
    don't post solutions, read the rules.
    My most unreserved apologies. I've never used the maths forum before, you see, and I thought I'd just try and help, but clearly the answer to -3/2 is too much of a spoiler to be said out loud.
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    (Original post by JackS94)
    My most unreserved apologies. I've never used the maths forum before, you see, and I thought I'd just try and help, but clearly the answer to -3/2 is too much of a spoiler to be said out loud.
    could you help me now then? ive posted the bit im not sure about
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    (Original post by LuckyEms)
    Thank you all!!

    Mind someone helping again?

    The next one is 9^x+1 = 3^x sqrt(27).

    I started by doing 3^x x 27^1/2 to get 81^x+1/2
    So i had 9^x+1=81^x+1/2

    Am i going in the right direction??
    is the question 9^x^+^1 =3^\sqrt[x]^(27)?
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    Take logs of both sides then use the power rule to place x in front of the log and then just solve that for x
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    (Original post by gaffer dean)
    is the question 9^x^+^1 =3^\sqrt[x]^(27)?
    No, the sqrt it seperate to the 3^x

    so its 3^x (multiplied by) sqrt(27)

    9^x^+^1 =3^x*\sqrt(27)
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    (Original post by LuckyEms)
    Thank you all!!

    Mind someone helping again?

    The next one is 9^x+1 = 3^x sqrt(27).

    I started by doing 3^x x 27^1/2 to get 81^x+1/2
    So i had 9^x+1=81^x+1/2

    Am i going in the right direction??
    don't do that (the bit i crossed out) you want everything to the power of 3 to work out what x is,
    3^x*\sqrt27

=3^x*(27)^1^/^2

=3^x*(3^3)^1^/^2
    (because 3^3=27) now use the rule (a^m)^n=a^m^n

    So, you would get 3^2^x^+^2=3^x * 3^3^/^2 = ...
    now you can work out the rest from here.
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    (Original post by LuckyEms)
    No, the sqrt it seperate to the 3^x

    so its 3^x (multiplied by) sqrt(27)

    9^x^+^1 =3^x*\sqrt(27)
    Oh okay, well some hints then. Root 27 is the same as (3^3)^(1/2). 9 is 3^2.
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    (Original post by JackS94)
    My most unreserved apologies. I've never used the maths forum before, you see, and I thought I'd just try and help, but clearly the answer to -3/2 is too much of a spoiler to be said out loud.
    That's ok.
    I'm sorry I didn't mean to be rude about it.
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    (Original post by LuckyEms)
    No, the sqrt it seperate to the 3^x

    so its 3^x (multiplied by) sqrt(27)

    9^x^+^1 =3^x*\sqrt(27)
    You want to try to get everything in terms of a common base, in this case 3.

    So for example 9 = 3^2 so you can rewrite the LHS as a power of 3.

    Similarly, on the RHS you can write 27 as a power of 3, and therefore you can write sqrt(27) as a power of 3.

    If you can rewrite your equation in the form
    3^a = 3^b
    where a and b are expressions that involve x, then you can write a = b and solve the resulting equation to get x.
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    Ok thank you everyone who helped!! I will have a go at this and let you know how i get on! Thank you!!
 
 
 
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